I posted this answer over at Math Stack Exchange, but since I'm not too sure about it I thought I should post it here as well. Hopefully someone who knows more about this than I do can look it over and check whether it's right.
I claim that the following statements hold for any subspace $W$ of $V$ (not just totally isotropic spaces). In the totally isotropic case, $W = R$ and $W^\perp = S$, which makes the solution slightly simpler.
Let $W$ be an arbitrary subspace of $V$, and let $G \leq O(V)$ be the group of orthogonal transformations that leave $W$ invariant. Define $$ R \;=\; W \cap W^\perp \qquad\text{and}\qquad S \;=\; W + W^\perp. $$ Note that $W^\perp$ is invariant under the action of $G$, and therefore $R$ and $S$ are also invariant.
The structure of $G$ is as follows. First, there is a short exact sequence $$ 1 \;\;\to\;\; A \;\;\to\;\; G \;\;\to\;\; O(S) \;\;\to\;\; 1 $$ where $A$ is an abelian group isomorphic to $K^d$ for some value of $d$. The homomorphism $G \to O(S)$ is surjective because of Witt's theorem.
The group $O(S)$ is a semidirect product. Specifically, $$ O(S) \;\cong\; \bigl( \text{Lin}(R,W/R) \times \text{Lin}(R,W^\perp/R) \bigr) \;\rtimes\; \bigl(O(R) \times O(W/R) \times O(W^\perp/R) \bigr). $$ Here $\text{Lin}(R,W/R)$ is the additive abelian group of all linear functions $R\to W/R$, and $\text{Lin}(R,W^\perp/R)$ is similarly an additive abelian group. Since $Q$ restricts to the null quadratic form on $R$, the orthogonal group $O(R)$ is the same as $GL(R)$. Moreover, since $Q$ is null on $R$, the quotients $W/R$ and $W^\perp/R$ are quadratic spaces, and $O(W/R)$ and $O(W^\perp/R)$ are the corresponding orthogonal groups. Note also that the quadratic forms on $W/R$ and $W^\perp/R$ are nondegenerate.
Edit: Here is a bit more information on the kernel $A$ of the epimorphism $G \to O(S)$. Since $Q|_{R\times S} = 0$, the quadratic form $Q$ defines a bilinear map $B \colon R \times (V/S) \to K$, and it is not hard to show that $B$ is a perfect pairing. It follows that the action of an element $g\in G$ on $V/S$ is entirely determined by the action of $g$ on $R$. In particular, every element of $A$ acts trivially on $V/S$. Therefore, every element $g\in A$ has the form $$ g(v) \;=\; v + \varphi(\pi(v)) $$ where $\pi\colon V \to V/S$ is the quotient map, and $\varphi\colon V/S \to S$ is a linear map. Thus $A$ is isomorphic to some subgroup of the abelian group $\text{Lin}(V/S,S)$.
To be specific, $A$ is isomorphic to the group of all linear maps $\varphi\colon V/S \to S$ that satisfy the following conditions:
The range of $\varphi$ lies in $R$.
The map $\varphi$ is "antisymmetric" with respect to $B$ in the sense that $$ B(\varphi(u),v) + B(\varphi(v),u) = 0 $$ for all $u,v \in V/S$.
In particular, $A$ is isomorphic to the additive group of all $m\times m$ antisymmetric matrices over $K$, where $m = \dim(R)$.
