Pick naturals $q,m,l,r$. Assume $q$ is a prime power and $2l\leq m$.
If $r\leq q^l+1$ then the maximal size of the union of $r$ many $(m-l)$-dimensional sub-vector spaces of $V=\mathbb{F}_q^m$ is $S=rq^{m-l}-(r-1)q^{m-2l}$. If $r\geq q^l+1$ then the size is $q^m$.

Note that for $r=q^l+1$ we get $S=q^m$. Since the maximal size of the union is obviously bounded from above by $q^m$ and is non-decreasing as a function of $r$,
it is enough then to show that for $r\leq q^l+1$ the maximal size of the union is $S$. Below we do assume $r\leq q^l+1$.

Assume having such $r$ subspaces with maximal union size.
Since the intersection of each two has dimension bounded from below by $m-2l$, the size of the intersection is bounded from above by $S$ and this is obtained iff they all have a common intersection of dimension $m-2l$.
In this case we may pass to the quotient space obtained by moding up this subspace, thus reduce to the special case $m=2l$ which is dealt by the following

**Lemma:** The maximal size of a collection of pairwise trivially intersecting $l$-dimensional subspace of $\mathbb{F}_q^{2l}$ is $q^l+1$.

An example of $q^l+1$ many such subspace is obtained by considerng the collection of lines inside a 2-dimensional space over $\mathbb{F}_{q^l}$.

On the other hand, if we have $s$ such subspaces and $s>q^l+1$ then the size of their union is $sq^l-(s-1)>q^{2l}$, which is an absurd.

**Bonus exercise:** based on the proof of the lemma above, prove that for general $l$ and $m$, the maximal number of pairwise trivially intersecting $l$-dimensional subspaces of $\mathbb{F}_q^m$ is $\frac{q^{\lfloor \frac{m}{l}\rfloor\cdot l}-1}{q^l-1}$.