Let $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ be a short exact sequence of groups. What can be said about group representations of $B$ if we assume a complete classification of the representations of $A$ and $C$? Precisely, assume we have determined the categories $Rep_R(A)$ and $Rep_R(C)$ of $R$-linear representations for a commutative ring $R$, up to equivalence.
If the sequence splits, then by identifying a representation of $B$ with its "classifying map" $B\rightarrow Aut(M)$ for an $R$-module $M$, we classify a representation of $B$ with its component representations of $A$ and $C$, so it seems like we have an equivalence between $Rep_R(B)$ and the pullback of the diagram $$ Rep_R(A) \rightarrow Mod_R \leftarrow Rep_R(C) $$ where both maps are the canonical forgetful functors taking a map $A \rightarrow Aut(M)$, say, to $M$.
What can be said if the sequence does not split?
I am most interested in the case when $B$ is a semidirect product of $A$ and $C$. I was inspired to ask this by this question, which asks about the specific case of the sequence $$0\rightarrow SO(n) \rightarrow O(n) \rightarrow \mathbb{Z}_2 \rightarrow 0$$
Even in the simplest case of a direct product, so that $B=A\times C$, this is complicated. The pullback of $$Rep_R(A) \rightarrow Mod_R \leftarrow Rep_R(C)$$ will give $R$-modules with actions of $A$ and $C$ that don't necessarily commute, so representations of the free product $A*C$ rather than the direct product $A\times C$.
If $R$ is a field of characteristic $p>2$ and $A$ and $C$ are cyclic groups of order $p$ then there's a fairly simple classification of modules for $RA$ and $RC$, but classifying $R[A\times C]$-modules is a "wild" problem that there's no hope of solving.