Any element $g$ in $\Omega_{2m+2}(F,Q)$ which fixes $v_1$ also stabilises $\langle v_1 \rangle^{\perp}$, so it has the following form with respect to $\mathcal{B}$: \begin{equation*} [g]_{\mathcal{B}}=\left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ u_2^\T &\ \ \ \ \ A\ \ \ \ \ & 0\ \\ & & \\ \hline \mu & u_1 & \lambda\ \end{array} \right], \end{equation*} where the matrix $A$ acts on the $2m$-dimensional subspace, spanned by $\{w_1, \ldots, w_{2m}\}$. Such an element $g$ acts on $v_2$ as \begin{equation*} \begin{array}{r@{\;}c@{\;}l} v_2 & \mapsto & [\mu \mid u_1 \mid\ \lambda], \end{array} \end{equation*} and since the bilinear form $f$ is preserved we get \begin{equation*} 1 = f(v_1,v_2) = f(v_1,\mu v_1) + f(v_1, [0 \mid u_1 \mid\ 0] ) + \lambda f(v_1,v_2) = \lambda. \end{equation*} Since $(v_1,v_2)$ is a hyperbolic pair, $f$ can be represented by the Gram matrix \begin{equation*} [f]_{\mathcal{B}} = \left[ \begin{array}{c|c|c} 0 & 0 & 1 \\ \hline & & \\ 0 &\ \ \ \ \ B\ \ \ \ \ & 0 \\ & & \\ \hline 1 & 0 & 0 \end{array} \right], \end{equation*} where $B$ is the matrix of $f_W$ with respect to the basis $\{ w_1 , \ldots, w_{2m} \}$. We explore the fact that an element in the stabiliser of $v_1$ preserves the form $f$: \begin{multline*} \left[ \begin{array}{c|c|c} 0 & 0 & 1 \\ \hline & & \\ 0 &\ \ \ \ \ B\ \ \ \ \ & 0 \\ & & \\ \hline 1 & 0 & 0 \end{array} \right] =\\ \end{multline*} \begin{multline*} = \left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ u_2^\T &\ \ \ \ \ A\ \ \ \ \ & 0\ \\ & & \\ \hline \mu & u_1 & 1\ \end{array} \right] \left[ \begin{array}{c|c|c} 0 & 0 & 1 \\ \hline & & \\ 0 &\ \ \ \ \ B\ \ \ \ \ & 0 \\ & & \\ \hline 1 & 0 & 0 \end{array} \right] \left[ \begin{array}{c|c|c} 1 & u_2 & \mu\ \\ \hline & & \\ 0 &\ \ \ \ \ A^\T\ \ \ \ \ & u_1^\T \\ & & \\ \hline 0 & 0 & 1\ \end{array} \right] = \end{multline*} \begin{equation*} = \left[ \begin{array}{c|c|c} 0 & 0 & 1\ \\ \hline & & \\ 0 &\ \ \ \ ABA^\T\ \ \ \ & ABu_1^\T + u_2^\T \\ & & \\ \hline 1 & u_2+u_1 BA^\T & 2\mu + u_1 B u_1^\T \end{array} \right], \end{equation*} so we notice that $ABA^\T = B$. Furthermore, since $[0\mid v \mid 0] [g]_{\mathcal{B}} = [0 \mid vA \mid 0]$, where $v \in W$, we obtain \begin{equation*} Q_W(vA) = Q([0 \mid vA \mid 0]) = Q([0 \mid v \mid 0]) = Q_W(v), \end{equation*} so $A$ is an element of $\GO_{2m}(F,Q)$. Additionally, $u_2 = -u_1BA^\T$ and we see that $u_2$ is uniquely determined by $u_1$. From the bottom right corner of the resulting matrix we obtain $\mu = -Q(u_1)$ in odd characteristic. In case of characteristic $2$ we can explore the quadratic form again: \begin{multline*} 0 = Q(v_2) = Q(v_2 [g]_{\mathcal{B}}) = Q( [\mu \mid u_1 \mid 1] ) = \\ = Q( [\mu \mid u_1 \mid 0] ) + Q(v_2) + f( [\mu\mid u_1 \mid 0], v_2) = Q(u_1) + \mu. \end{multline*} Consider the decomposition \begin{equation*} \left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ -ABu_1^\T &\ \ \ \ \ A\ \ \ \ \ & 0\ \\ & & \\ \hline -Q(u_1) & u_1 & 1 \end{array} \right] = \left[ \begin{array}{c|c|c} 1 & 0 & 0 \\ \hline & & \\ 0 &\ \ \ \ \ A\ \ \ \ \ & 0 \\ & & \\ \hline 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ -Bu_1^\T &\ \ \ \ \ \II_{2m}\ \ \ \ \ & 0\ \\ & & \\ \hline -Q(u_1) & u_1 & 1 \end{array} \right]. \end{equation*} The matrices of the form \begin{equation*} C_{u_1} = \left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ -Bu_1^\T &\ \ \ \ \ \II_{2m}\ \ \ \ \ & 0\ \\ & & \\ \hline -Q(u_1) & u_1 & 1 \end{array} \right] \end{equation*} generate an elementary abelian group isomorphic to $W$ (as abelian groups). Indeed, since the product of two such matrices is given by \begin{multline*} \left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ -Bu^\T &\ \ \ \ \ \II_{2m}\ \ \ \ \ & 0\ \\ & & \\ \hline -Q(u) & u & 1 \end{array} \right]\left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ -Bv^\T &\ \ \ \ \ \II_{2m}\ \ \ \ \ & 0\ \\ & & \\ \hline -Q(v) & v & 1 \end{array} \right]= \\ =\left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ -B(u+v)^\T &\ \ \ \ \ \II_{2m}\ \ \ \ \ & 0\ \\ & & \\ \hline -Q(u+v) & u+v & 1 \end{array} \right], \end{multline*} we see that the set of these matrices is closed under multiplication and moreover any two such matrices commute. To show that the matrix $A$ is an element of $\Omega_{2m}(F,Q)$, we use Proposition 1.6.11 from [BHRD] to calculate the spinor norm and, in case of characteristic $2$, the quasideterminant of the matrices $C_{u_1}$. Note that $\det(C_{u_1}) = \det([g]_{\mathcal{B}}) = 1$. Consider the matrix \begin{equation*} \II-C_{u_1} = \left[ \begin{array}{c|c|c} 0 & 0 & 0 \\ \hline & & \\ Bu_1^\T &\ \ \ \ \ 0\ \ \ \ \ & 0\ \\ & & \\ \hline Q(u_1) & -u_1 & 0 \end{array} \right]. \end{equation*} For a vector $v$ we denote by $[v]_i$ its $i$-th component. Now, if $u_1 = 0$, then $\II-C_{u_1}$ has rank $0$, whereas if $u_1 \neq 0$, then there is an index $i$ such that $[Bu_1^\T]_i \neq 0$ and it follows that the rank of $\II-C_{u_1}$ in this case is $2$. Consequently, $k=\rank(\II-C_{u_1})$ is even, and so by the Proposition 1.6.11 in [BHRD], the quasideterminant of $C_{u_1}$ is $1$. Further, if $D$ is a $k\times (2m+2)$ matrix whose rows are the basis elements of a complement of the nullspace of $\II-C_{u_1}$, then the spinor norm of $C_{u_1}$ is $1$ if $\det(D(\II-C_{u_1}) [f]_{\mathcal{B}} D^\T)$ is a square in $F$. If $u_1 \neq 0$, then the complement of the nullspace of $I-C_{u_1}$ has the basis $\{ w_i, v_1 \}$, where the index $i$ is such that $[Bu_1^\T]_i \neq 0$. The matrix $D$ can be taken to have the following form: \begin{equation*} D = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix}, \end{equation*} where $1$ in the first row is in the $(i+1)$-st position. We calculate \begin{equation*} D(\II-C_{u_1}) = \left[ \begin{array}{c|c|c} \alpha & 0 & 0 \\ \hline Q(u_1) & -u_1 & 0 \end{array} \right],\ [f]_{\mathcal{B}} D^\T = \begin{bmatrix} 0 & 1\ \\ B_{1,i} & 0\ \\ \vdots & \vdots \\ B_{2m,i} & 0\ \\ 0 & 0 \end{bmatrix}, \end{equation*} where $\alpha = [Bu_1^\T]_i = [u_1B]_i$. Finally, \begin{equation*} D(\II-C_{u_1}) [f]_{\mathcal{B}} D^\T = \begin{bmatrix} 0 & \alpha \\ -\alpha & Q(u_1) \end{bmatrix}, \end{equation*} so $\det(D(\II-C_{u_1})FD^\T) = \alpha^2$ as needed. Since the quasideterminant and the spinor norm are multiplicative (Theorems 11.43 and 11.50 in [Taylor]), and $g \in \Omega_{2m+2}(F,\hat{Q})$, we conclude that $A$ acts on $W$ as an element of $\Omega_{2m}(F,Q)$ and it follows that the stabiliser of $v_1$ in $\Omega_{2m}(F,\hat{Q})$ is indeed a subgroup of shape $W\cn\Omega_{2m}(F,Q)$. Lastly, if we stabilise $v_1$ and $v_2$ simultaneously, a general element in the stabiliser takes the form \begin{equation*} \left[ \begin{array}{c|c|c} 1 & 0 & 0 \\ \hline & & \\ 0 &\ \ \ \ \ A\ \ \ \ \ & 0 \\ & & \\ \hline 0 & 0 & 1 \end{array} \right], \end{equation*} so the stabiliser of the pair $(v_1,v_2)$ is $\Omega_{2m}(F,Q)$.

Suppose $u, v \in O_{\lambda}$. Since by Witt's lemma $\GO_{2m+2}(F,Q)$ acts transitively on $O_{\lambda}$, there is an element $g \in \GO_{2m+2}(F,Q)$ which sends $u$ to $v$. Consider a basis for our vector space $V$, $\mathcal{B} = \{v_1, w_1, \ldots, w_{2m-2}, v_1\}$, such that as before $(v_1,v_2)$ is a hyperbolic pair and $u \in \langle v_1, v_2 \rangle$, and so $V = \langle v_1 \rangle \oplus \langle w_1, \ldots, w_{2m-2} \rangle \oplus \langle v_2 \rangle$. Suppose $h$ is an element in the stabiliser of $(v_1,v_2)$. As a consequence, $h$ stabilises $u$ and with respect to $\mathcal{B}$ it has the form \begin{equation*} [h]_{\mathcal{B}} = \left[ \begin{array}{c|c|c} 1 & 0 & 0 \\ \hline & & \\ 0 &\ \ \ \ \ A\ \ \ \ \ & 0 \\ & & \\ \hline 0 & 0 & 1 \end{array} \right], \end{equation*} where the matrix $A$ acts on $W=\langle w_1, \ldots, w_{2m-2} \rangle$ as an element of $\GO_{2m}(W,Q_W)$. If the determinant of $g$ is $1$, then we may take \begin{equation*} A = \begin{bmatrix} \mu & 0 & 0 & \cdots & 0 \\ 0 & \mu^{-1} & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}, \end{equation*} where $\mu \in F$. On the other hand, if $\det(g) = -1$, then we take \begin{equation*} A = \begin{bmatrix} 0 & \mu^{-1} & 0 & \cdots & 0 \\ \mu & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} \end{equation*} instead, so we can always get $\det(hg) = 1$. Note that the latter choice of $A$ also adjusts the quasideterminant in characteristic $2$ as the rank of $\II-[h]_{\mathcal{B}}$ in this case is odd. Finally, choosing $\mu$ accordingly we can ensure that the spinor norm of $hg$ is $1$, i.e. $hg \in \Omega_{2m+2}(F,Q)$.

Recall that $v_2 \in V$ is mapped under the action of $\mathcal{G} = W\cn\Omega_{2m}(F,Q_W)$ to a vector of the form $[-Q_W(u) \mid u \mid 1]$, where $u \in W$. Since the stabiliser of $v_2$ in $\mathcal{G}$ is $\Omega_{2m}(F,Q_W)$, we conclude that the orbit of $v_2$ under the action of $\mathcal{G}$ is the following set: \begin{equation*} \Orb_{\mathcal{G}}(v_2) = \left\{\ \left[ -Q_W(u) \mid u \mid 1 \right]\ \big|\ u \in W \ \right\}. \end{equation*} Since the elements of this orbit are in one-to-one correspondence with the cosets of $\Omega_{2m}(F,Q_W)$ in $\mathcal{G}$, it is enough to show the primitive action on $\Orb_{\mathcal{G}}(v_2)$. Consider the action of $\mathcal{G}$ on $\Orb_{\mathcal{G}}(v_2)$. A general element in $\mathcal{G}$ acts on the elements of $\Orb_{\mathcal{G}}(v_2)$ in the following way: \begin{multline*} [ -Q_W(u) \mid u \mid 1 ] \ \mapsto \ [ -Q_W(u) - uABv^\T - Q(v) \mid uA + v \mid 1] = \\ = [-Q_W(uA + v) \mid uA + v \mid 1]. \end{multline*} Note that $uABv^\T = f_W( uA, v )$. We see that this action is isomorphic to the action on $W$ defined by $u \mapsto uA + v$, where $u,v \in W$. In case when $A$ is the identity matrix, this map is a translation. On the other hand, taking $v = 0$, we obtain the action of $\Omega_{2m}(F,Q_W)$. Denote the group generated by the described action on $W$ as $\mathrm{A\Omega}_{2m}(F,Q_W)$. Since $Q_W$ is of Witt index at least $1$, we may choose a hyperbolic pair $(u_1,u_2)$ in $W$ such that $W = \langle u_1, u_2 \rangle \oplus U$, where $U = \langle u_1, u_2 \rangle^{\perp}$. We aim to show that any $\mathrm{A\Omega}_{2m}(F,Q_W)$-congruence on $W$ is trivial (and hence, the action is primitive). Suppose $w_1 \sim w_2$, where $w_1,w_2 \in W$ and $\sim$ is some congruence relation preserved by $\mathrm{A\Omega}_{2m}(F,Q_W)$. It follows that $w_1 - w_2 \sim 0$, and so we may start with $v \sim 0$ for some $v \in W$. We distinguish two cases. First, if $Q_W(v) = 0$, then since $\Omega_{2m}(F,Q_W)$ acts transitively on isotropic vectors in $W$, we get $u_1 \sim 0$, $u_2 \sim 0$, and also $-\lambda u_2 \sim 0$ for any $\lambda \in F$. Since $\sim$ is a congruence relation, it is transitive and so $u_1 \sim -\lambda u_2$, from which it follows $u_1 + \lambda u_2 \sim 0$. Now, $Q_W(u_1 + \lambda u_2) = \lambda$, and $\lambda$ is an arbitrary field element, so $\sim$ is trivial. Next, if $Q_W(v) = \lambda$ for some non-zero $\lambda \in F$, we consider two vectors $w_1 = u_1 + \lambda u_2$ and $w_2 = u_1 + (\lambda - Q_W(u))u_2 + u$ for some $u \in U$. Note that $Q_W(w_1) = Q_W(w_2) = \lambda$, so since $\Omega_{2m}(F,Q_W)$ acts transitively on the vectors of norm $\lambda$, we obtain $w_1 \sim 0$ and $w_2 \sim 0$, from which it immediately follows that $w_1 \sim w_2$, and further $w_1 - w_2 \sim 0$. We find $Q_W(w_1 - w_2) = Q_W(u - Q(u)u_2) = Q_W(u)$, so in fact we have $u \sim 0$ for some $u \in U$. From $Q(u_1+u) = Q(u)$ it follows that $u_1 + u \sim 0$, and so by transitivity $u_1+u \sim u$. We subtract $u$ from both sides to obtain $u_1 \sim 0$, which is covered by the previous case.

Let $G = \Omega_{2m+2}(F,Q)$. We aim to prove that if $v_1 \sim v$ for some $v \in V$, where $\sim$ is any non-trivial $G$-congruence, then $v = \lambda v_1$ for some $\lambda \in F$. For the sake of finding a contradiction, suppose that $v = \lambda v_1 + u + \mu v_2$, where $u + \mu v_2 \neq 0$, i.e. either $u \neq 0$ or $\mu \neq 0$. We distinguish two cases. First, if $\mu = 0$, then $v_1 \sim v$, where $v = \lambda v_1 + u$ with $0\neq u \in W$. We have $0 = Q(v) = Q_W(u)$. Recall that $W\cn \Omega_{2m}(F,Q_W)$ is the stabiliser in $\Omega_{2m+2}(F, Q)$ of $v_1$, so with respect to the familiar basis $\mathcal{B} = \{v_1, w_1 ,\ldots, w_{2m}, v_2\}$, its general element has the form \begin{equation*} [g]_{\mathcal{B}}=\left[ \begin{array}{c|c|c} 1 & 0 & 0\ \\ \hline & & \\ u_2^\T &\ \ \ \ \ A\ \ \ \ \ & 0\ \\ & & \\ \hline \nu & u_1 & 1\ \end{array} \right], \end{equation*} where $u_2 = -u_1 BA^{\T}$. Now, $g$ maps $v$ to a vector of the form $v^g = (\lambda + u u_2^{\T}) v_1 + uA$. Since $u u_2^{\T} = -u A B u_1^{\T} = - f_W(u, u_1)$, we get $v^g = (\lambda - f_W(u, u_1))v_1 + uA$. Setting $A = \II_{2m}$, we see that it is possible to send $v$ to a vector of the form $\alpha v_1 + u$ for any $\alpha \in F$. On the other hand, taking $u_1 = 0$, we obtain the action on $W = \langle w_1, \ldots, w_{2m} \rangle$ of $\Omega_{2m}(F,Q_W)$, which is transitive on the isotropic vectors in $W$. We have shown that $v_1 \sim v$ for any $v$ of the form $\alpha v_1 + u$ for arbitrary $\alpha \in F$ and $0\neq u \in W$. The group $\Omega_{2m+2}(F,Q)$ in its turn is transitive on the isotropic vectors in $V$, so there exists an element $h \in \Omega_{2m+2}(F,Q)$ such that $v_1^h = v_2$ and $v^h \in W \oplus \langle v_2 \rangle$. It follows that $v_2 \sim v^h$ and so, by transitivity of $\sim$ we find that $v_1 \sim v_2$. Finally, it is easy to derive the congruences of the form $v_1 \sim \beta v_1$ and $v_2 \sim \beta v_2$ for any non-zero $\beta \in F$. Indeed, there exists an element $g \in \Omega_{2m+2}(F,Q)$ such that $v_1^g = \beta v_1$ for some non-zero $\beta \in F$. We then have $\beta v_2 \sim \alpha \beta v_1 + u^g$, so $v_1 \sim \beta v_1$. Similarly we obtain $v_2 \sim \beta v_2$ for any non-zero $\beta \in F$, and so $\sim$ is universal, contradiction.

The case $\mu = 0$ can be established with essentially similar reasoning.