If the dimension of $V$ is even, $2n$, then there are two families of totally
isotropic subspaces of dimension $n$ (if the form is split otherwise there may
be no such subspace at all). Two such subspaces $W$ and $W'$ belong to the same
family precisely when the parity of $\dim W\cap W'$ is the same as that of
$n$. The two families are stable under $\mathrm{SO}(q)$ (and permuted by the
rest of $\mathrm O(q)$). There certainly is an elementary proof of this fact yet
I think that the "real" reason comes from algebraic geometry. The two families
are disjoin algebraic subvarieties of the Grassmannian of $n$-spaces in $V$ and
$\mathrm{SO}(q)$ is connected and hence must preserve them.

In any case when $n=1$ everything is very elementary; we can choose an isotropic
basis $e_1,e_2$ with $e_1\cdot e_2=1$ and then $\mathrm{SO}(q)$ consists of the
diagonal matrices of determinant $1$ which hence fixes the two isotropic
subspaces (spanned by $e_1$ and $e_2$ respectively) while the orthogonal
matrices of determinant $-1$ permute them.

In all other case $\mathrm{SO}(q)$ acts transitively on the totally isotropic spaces of fixed dimension (which is easily proven).