Let $X\xrightarrow{f}Y\xrightarrow{q}Z$ be morphisms of schemes such that $f$ is finite and $q$, $q\circ f$ are compactifiable, which means they can be factored as an open immersion followed by a proper morphism. Let $\cal{F}$ be a torsion étale sheaf on $X$. Is it true that there is an isomorphism \begin{equation} R^r(q\circ f)_!{\cal{F}} \cong R^rq_!(f_*\cal{F}) \end{equation} of higher direct images with proper support?
Let me say a bit of why I suspect this could be true. Since $f$ is finite $f_*$ is exact, and it also preserves injectives because its left adjoint $f^*$ is exact. Therefore if we take higher direct images instead, we do have an isomorphism $R^r(q\circ f)_*{\cal{F}} \cong R^rq_*(f_*\cal{F})$. This can be seen directly by the definition of derived functors with injective resolutions. But $R^rq_!$ is not a derived functor, but rather is defined as $R^rq_!:=(R^r\bar{q}_*)\circ j_!$ where $q=\bar{q}\circ j$ is a compactification and $j_!$ denotes extension by zero.
I have been able to prove the following similar result: if $X\xrightarrow{p}Y\xrightarrow{g}Z$ with $g$ finite and $p,g\circ p$ compactifiable, then \begin{equation}R^r(g\circ p)_!{\cal{F}}\cong g_*R^rp_!\cal{F}.\end{equation} The key is noting that if $p=\bar{p}\circ j$ is a compactification of $p$, then $g\circ p=g\circ\bar{p}\circ j$ is a compactification of $g\circ p$ because $g$ is proper, and thus $R^r(g\circ\bar{p})_*(j_!{\cal{F}})\cong g_*R^r\bar{p}_*(j_!\cal{F})$ because $g_*$ is exact. However for my question above I'm just not able to find compactifications of $q$ and $q\circ f$ which are related to each other, or know if this even is a good approach.