= Solution
Keep row-vector action and the block order $(W',U,W)$. Let $R$ be the reversal matrix of size $k$, arising from the original reversed order of $f_1,\ldots,f_k$, and let $J_U$ be the <alternating bilinear form> matrix on $U$. Then
$$
J=\begin{pmatrix}0&0&R\\0&J_U&0\\-R&0&0\end{pmatrix},\qquad R=R^T=R^{-1}.
$$
For an element of $Q$, the equation $gJg^T=J$ gives
$$
E=J_UD^TR,\qquad FR-(FR)^T=DJ_UD^T.
$$
Thus $D$, a $k\times(2m-2k)$ matrix, is arbitrary and uniquely determines $E$. The right side of the second equation is alternating, including in characteristic two: its diagonal entries vanish because $J_U$ represents an <alternating bilinear form>.
For any alternating matrix $K$, the equation $T-T^T=K$ has exactly $q^{k(k+1)/2}$ solutions. For each pair $i<j$, choose one entry freely and solve for the opposite entry; each diagonal entry is free. This works in characteristic two as well as odd characteristic. Taking $T=FR$ therefore gives
$$
\boxed{|Q|=q^{2k(m-k)}q^{k(k+1)/2}.}
$$
This is the <unipotent radical count for a symplectic parabolic subgroup>. It does not incorrectly replace the alternating constraint by division by two.
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