If we have $H$ and $G$ that acts on a set $X$ on right and left, this is $h.x\in X$ and $x.g\in X$ (with $h\in H$ and $g\in G$) that are compatible, this is $h.(x.g)=(h.x).g$ with $g\in G$, $h\in H$ and $x\in X$.
So if we pick an $x\in X$ what can we say about: $\{(g,h)\in G\times H | h.x.g=x\}$
It contains the product of stabilizers $S_G(x)\times S_H(x)$ but we have the equality. If not how can we write the set above?
(We may assume that $H$ and $G$ and $X$ are all finite.)
Or more concretely we may assume $H$ and $G$ are matrix groups over a finite field (the same) inside some $U_n(\mathbb{F}_q)$. And $X$ we may assue that is an subalgebra of the unitriangular algebra (the matrix algebra consisting of matrix strictly upper diagonal), and the action is given by matrix multiplication. (where $U_n(\mathbb{F}_q)$ is the unitriangular group)
One example to show that equality may not hold in general: Let $G = H$ be an abelian group acting on itself by $$ h\cdot x:= hx \text{ and } x\cdot g := g^{-1}x $$ Then for any $x\in G$, $$ S_G(x) = \{g\in G : gx = x\} = \{e\} = S_H(x) $$ However, $$ g\cdot x\cdot g = x \quad\forall g\in G $$ and so the stabilizer you have written down contains the diagonal set $$ \{(g,g) : g\in G\} $$
As for how to write this set in general, I am not sure. If you have a specific example in mind, then do include that in the question, and perhaps that might have a meaningful description.