I posted this answer over at [Math Stack Exchange][1], 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][2].

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:

1. The range of $\varphi$ lies in $R$.

2. 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 $m = \dim(R)$.


  [1]: https://math.stackexchange.com/questions/43500/subgroup-of-oq
  [2]: http://en.wikipedia.org/wiki/Witt%27s_theorem