Your general question seems too general. Here is a partial answer to your specific question.

Let $V = \mathbb{F}^{2m}$ and $G = \mathrm{Sp}_{2m}(\mathbb{F})$. Consider some $vg \in VG$. We know that $g = [x,y]$ for some $x, y \in G$. Assume $y$ can be chosen so that $y-1$ is invertible. (Can this always be done?) Now observe that $$[wx,y] = x^{-1} w^{-1} w^y x^y = (-wx + wyx) [x,y]\qquad (w \in V).$$ Since $y-1$ is invertible, so is $yx-x$, so we can choose $w$ so that $-wx+wyx = v$ and hence $vg = [wx,y]$.

Your general question seems too general. Here is a partial answer to your specific question.

Let $V = \mathbb{F}^{2m}$ and $G = \mathrm{Sp}_{2m}(\mathbb{F})$. Consider some $vg \in VG$. We know that $g = [x,y]$ for some $x, y \in G$. Assume $y$ can be chosen so that $y-1$ is invertible. Now observe that $$[wx,y] = x^{-1} w^{-1} w^y x^y = (-wx + wyx) [x,y]\qquad (w \in V).$$ Since $y-1$ is invertible, so is $yx-x$, so we can choose $w$ so that $-wx+wyx = v$ and hence $vg = [wx,y]$.

I think such a $y$ probably can be chosen for any given $g \in G$. If $|\mathbb F| \equiv 1 \pmod 4$, this follows from a result of Gow [1], which states that $G = C^2$ where $C$ is the single conjugacy class consists of elements $y$ such that $y^2 = -1$ (verifying Thompson's conjecture in this case). Such $y$ clearly do not have eigenvalue $1$, so $y-1$ is invertible as required, and since $C^2 = G$ there must be some $y \in C$ and $x \in G$ such that $g = y^{-x}y = [x,y]$.

[1] Gow, Roderick, Commutators in the symplectic group, Arch. Math. 50, No. 3, 204-209 (1988). ZBL0628.20037.