This is more easily understood as a question about alternating tensors. The space of skew-symmetric $m\times m$ matrices over a field $k$ is isomorphic to $\Lambda^2(k^m)$, via $u\wedge v\mapsto uv^T-vu^T$. Then a matrix is of rank $\leq 2$ if and only if it is a pure alternating tensor $u\wedge v$. Now it is easy to check that a vector space of skew-symmetric matrices consists only of matrices of rank $\leq 2$ if and only if it is of the form $e_1\wedge U$ where $U$ is a subspace of $k^m$. Let us fix once and for all a splitting $k^m=ke_1+W$. We can clearly assume that $U\subset W$.
A skew-symmetric matrix $Q$ has rank $4$ if and only if it is of the form $a\wedge b+c\wedge d$ where $a,b,c,d$ are linearly independent. Now if $Q-e_1\wedge e_2$ has rank 2 then $$a\wedge b+c\wedge d=e_1\wedge e_2+e_3\wedge e_4$$ for some $e_3,e_4$ and $a,b,c,d$ span the same subspace as $e_1,e_2,e_3,e_4$. More precisely, these expressions are equal if and only if the matrix with columns $a,b,c,d$, expressed in terms of $e_1,e_2,e_3,e_4$ is a symplectic $4\times 4$ matrix (where the symplectic form is determined by $e_1\wedge e_2+e_3\wedge e_4$). After applying a suitable symplectic transformation, we may assume that $a=e_1$ and $b,c,d,e_2,e_3,e_4\in W$.
So the problem has been reduced to the following: given $b,c,d\in W$ linearly independent, how many $u\in U$ are there such that $e_1\wedge b+c\wedge d+e_1\wedge u$ has rank 2?
Obviously this is equivalent to $e_1,(b+u),c,d$ being linearly dependent. But this is equivalent to $b+u,c,d$ being linearly dependent, so we obtain our answer. The set of all such $u$ is given as the intersection $$(-b+kc+kd)\cap U.$$
In the case of a finite field $k={\mathbb F}_q$, this set consists of at most $q^2$ elements, and that happens precisely when $b,c,d\in U$.
