Akhil, you're thinking of this the opposite of how I think group cohomology was discovered. The concept of group cohomology originally centered around the questions about the (co)homology of $K(\pi,1)$-spaces, by people like Hopf (he called them aspherical rather than $K(\pi,1)$ spaces, and Hopf preferred homology to cohomology at that point). I think the story went that Hopf observed his formula for $H_2$ of a $K(\pi,1)$, which was a description of $H_2$ entirely in terms of the fundamental group of the space.
This motivated people to ask to what extent (co)homology is an invariant of the fundamental group of a $K(\pi,1)$-space. This was resolved by Eilenberg and Maclane. Eilenberg and Maclane went the extra step to show that one can define cohomology of a group directly in terms of a group via what nowadays would be called a "bar construction" (ie skipping the construction of the associated $K(\pi,1)$-space). Bar constructions exist topologically and algebraically and they all have a similar feel to them. On the level of spaces, bar constructions are ways of constructing classifying spaces. For groups they construct the cohomology groups of a group. The latter follows from the former -- if you're comfortable with the concept of the "nerve of a category", this is how you construct an associated simplicial complex to a group (a group being a category with one object). The simplicial (co)homology of this object is your group (co)homology.
Dieudonne's "History of Algebraic and Differential Topology" covers this in sections V.1.D and V.3.B. I don't think that answers all your questions but it answers some.
Consider the simplical set $EG$. It has a free $G$-action and is contractible, so that $C_*(\mathbb Z[EG])$ is a complex of free $\mathbb Z[G]$-modules with a quasi-isomorphism $C_*(\mathbb Z[EG]) \to \mathbb Z[0]$.
$EG$ in degree $n$ is $G^{n+1}$ so this chain complex is $\mathbb Z[G^{n+1}]$ in degree $n$, and the differential is $\sum_i (-1)^i d_i$.
($d_i$ is the one I mentioned in the question, and the $G$-action is coordinate-wise by translation)
The point is that the writing $\mathbb Z[G^{n+1}]$ doesn't make the free-ness apparent, and one way to do so will introduce the $G$-action on $d_0$.
Namely, if $X$ is an arbitrary $G$-set, then $G\times X^{triv}\cong G\times X$ via $(g,x)\mapsto (g,gx)$ (this is obviously a bijection, and it is a simple computation to check that it is $G$-equivariant). In particular, $G^{n+1}\cong G\times (G^n)^{triv}$ via $\varphi: (g,x)\mapsto (g, g^{-1}x)$. One issue with this is that it doesn't respect the simplicial structure, but of course, it shouldn't, and can't ! $EG$ is not $G\times -$ as a $G$-simplicial set. A bigger issue is that the formula for $d_0$ we get is not that good if we do this. But luckily, in this case, we can trivialize the $G^n$ part in several steps.
Indeed, $G^{n+1} = G^{n-1}\times G\times G\cong G^{n-1}\times G\times G^{triv}$, and then you can again remove one $G$ by doing $G\times G\times G^{triv}\cong G\times (G\times G^{triv})^{triv}$, and so on and so forth. If you work this out, you'll see that the isomorphism is given, on $G^{n+1}$, by $G\times (G^n)^{triv}\cong G^{n+1}, (g_0,...,g_n)\mapsto (g_0,g_0g_1,g_0g_1g_2,...., g_0...g_n)$ (you can check a posteriori that this is a $G$-map, and a bijection). Call this $\varphi^{-1}$.
If we compute what $d_i$ looks like with this presentation, we see that for $0\leq i< n$, $\varphi \circ d_i \circ \varphi^{-1} (g_0,...,g_n) =\varphi\circ d_i (g_0, g_0g_1,...,g_0...g_n) = \varphi (g_0,g_0g_1, ...,\widehat{g_0...g_i} ,..., g_0...g_n) = (g_0,...,g_ig_{i+1},...,g_n)$
For $i=n$, $\varphi \circ d_n \circ \varphi^{-1} (g_0,...,g_n)= \varphi\circ d_n(g_0,g_0g_1,...,g_0...g_n) = \varphi(g_0,g_0g_1,...,g_0...g_{n-1}) = (g_0,...,g_{n-1})$.
The important point is to observe that for $i=0$, this does not send $\{e\}\times G^n$ to $\{e\}\times G^{n-1}$
So if we write the $n$-term of our chain complex as $\mathbb Z[G]\otimes_\mathbb Z\mathbb Z[(G^n)^{triv}]$ to witness that it is indeed a free $G$-abelian group, and compute the differential on $\hom_G(\mathbb Z[G]\otimes_\mathbb Z[G^n],A)$, we see that if we view elements of this as set maps $G^n\to A$, then for all differentials except for $d_0$, we have nothing to do, but precomposition by $d_0$ makes us go out of $G^n$ so there is something to say.
Precomposition by $d_0$ looks like this : if $f: G^n\to A$, then we can view $f$ as a $G$-map $\tilde f: G\times G^n\to A$ by sending $(g_0,...,g_n)$ to $g_0\cdot f(g_1,..,g_n)$.
So now, the map $G^{n+1}\to A$ associated to $\tilde f\circ d_0$ is defined by $\tilde f\circ d_0 (e,...,g_{n+1}) $, and this is $\tilde f(g_1,..., g_{n+1}) = g_1\tilde f(e,g_2,...,g_{n+1}) =g_1 f(g_2,...,g_{n+1})$.
So this gives us the desired $d_0$, where the other $d_i$'s preserve $\{e\}\times G^n$, so there is nothing to say.
Best Answer
You're skipping the step where you go from the free resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ to actually computing the Ext. The free resolution is the cellular chain complex of $\widetilde{Y}$. However, to compute $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$, you have to Hom this resolution into $A$ before taking cohomology. If $F$ is a free $\mathbb{Z}G$-module with basis $B$, then $\operatorname{Hom}_{\mathbb{Z}G}(F,A)$ is just $A^B$. Now, if $C_n(\widetilde{Y})$ denotes the $n$th cellular chain group of $\widetilde{Y}$, the $n$-cells of $\widetilde{Y}$ form a basis for $C_n(\widetilde{Y})$ over $\mathbb{Z}$. But over $\mathbb{Z}G$, a basis would instead be a set of representatives of the $G$-orbits of the $n$-cells of $\widetilde{Y}$. The $G$-orbits of the $n$-cells of $\widetilde{Y}$ just correspond to the $n$-cells of $Y$, so you can identify a basis for $C_n(\widetilde{Y})$ with the set of $n$-cells of $Y$. This means that when you Hom into $A$ over $\mathbb{Z}G$, you just get the group of functions from the set of $n$-cells of $Y$ to $A$, which is exactly the $n$th cellular cochain group of $Y$ with coefficients in $A$. So, the chain complex that computes $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$ has the same groups in it as the cellular cochain complex of $Y$ with coefficients in $A$. (Of course, you have to do some additional work to check that the differentials in these chain complexes are also the same.)