I am familiar with two definitions of group cohomology:
The group cohomology of $G$ is the (singular) cohomology of the classifying space $BG$.
Let $M$ be a $G$-module (an abelian group with a $G$-action). Let $C^n(G,M) = \{f:G^n\to M\}$, and define $d^n:C^n(G,M) \to C^{n+1}(G,M)$ by $$(d^n f)(g_1,\ldots,g_{n+1}) = g_1\cdot f(g_2,\ldots,g_{n+1}) + \sum_{i = 1}^n (-1)^i f(g_1,\ldots,g_{i-1},g_i g_{i+1},\ldots, g_{n+1}) + (-1)^{n+1}f(g_1,\ldots,g_n).$$
We then define the cohomology of $G$ to be the homology of this chain complex.
Nominally, these two definitions must give different objects, as one starts with additional data of a $G$-module, $M$, and the other does not. I would like to reconcile these two definitions.
When $M$ carries the trivial $G$ action, I can show that the two definitions are identical by tracing through the construction of $BG$, computing face maps, and thereby the differential operator of simplicial cohomology. I will now outline this procedure.
$BG$ is the quotient of $EG =${$ (g_1,\ldots,g_n) \mid n \in \mathbb{N}, g_i \in G $} (where each tuple is thought of as a simplex) by the diagonal action $g\cdot (g_1,\ldots,g_n) = (gg_1,\ldots, gg_n)$. An n-simplex in $BG = EG/G$ is an equivalence class of tuples, which can be uniquely represented by $n$ elements of $G$ by considering the ratios between adjacent elements in the tuple. Note that this is not done by normalizing the first coordinate to be $e$—the ratios $[g_1\mid g_2\mid \cdots \mid g_n]$ represent the equivalence class of the tuple $(g_1,g_1g_2,g_1g_2g_3,\ldots, g_1\cdots g_n)$.
We can then compute the face maps using these coordinates. The first and last face maps simply drop the first or last coordinate, and the $i^{\text{th}}$ face of $[g_1\mid g_2\mid \cdots \mid g_n]$ is the $(n-1)$-simplex $[g_1\mid \cdots g_{i-1} \mid g_ig_{i+1}\mid g_{i+2}\mid \cdots g_n]$. We can then plug this into the standard formula for the boundary map in (simplicial) cohomology to obtain
$$(d^n f)(g_1,\ldots,g_{n+1}) = f(g_2,\ldots,g_{n+1}) + \sum_{i = 1}^n (-1)^i f(g_1,\ldots,g_{i-1},g_i g_{i+1},\ldots, g_{n+1}) + (-1)^{n+1}f(g_1,\ldots,g_n),$$
where we now interpret $(g_1,\ldots,g_{n+1})$ as a simplex in $BG$ via the ratio-coordinates, and recognize each term in the above expression as the evaluation of $f$ on a face of the simplex $(g_1,\ldots,g_{n+1})$.
This is a satisfying derivation in that it explains the funny-looking definition of the boundary map in the chain map described above. It is unsatisfying in that it only works when $M$ carries the trivial action—the first term in the definition is $g_1\cdot f(g_2,\ldots,g_{n+1})$, while in my classifying space story, it is $f(g_2,\ldots,g_{n+1})$.
Question: is there a(n ideally, topological) way to understand what the group action is doing here? Why does $g_1\cdot $ shows up in the definition of the boundary map?
I would prefer an answer that does not contain the phrase "derived functor".