All right, here's a solution that ought to work for any algebraically closed field of characteristic $0$ in the case where $Q$ is nondegenerate. This is all based on a bunch of matrix calculations, so I don't have proofs for some of it, but it ought to be correct unless I've made a calculation mistake. The obvious way to check it would be to prove these statements rigorously.
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.
As for the question of connectivity: since the kernels $A$ and $\text{Lin}(R,W/R)\times\text{Lin}(R,W^\perp/R)$ are both connected, the group $G$ will be connected if and only if the groups $O(R)$, $O(W/R)$, and $O(W^\perp/R)$ are connected. Note that $O(R) \cong GL(R)$, while $O(W/R)$ and $O(W^\perp/R)$ are orthogonal groups for nondegenerate quadratic spaces. (Perhaps it is obvious to you whether these are connected -- I do not know much about the connectedness of orthogonal groups over arbitrary fields.)
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 $m = \dim(R)$.