Generally looks good, though as Berci says in the comments you should check that $\phi$ doesn't depend on your choices.
However, I write an answer, because I would suggest an alternative method of proof for
$\newcommand\im{\operatorname{im}}\ker\alpha^*=\im\beta^*$.
The key is to notice that $C\cong B/\im\alpha$, and $\beta : B\to C$ is the quotient map.
Therefore you can use the universal property of the quotient, which is that maps $\psi : B\to M$ such that $\psi(\im\alpha)=0$ are in one-to-one correspondence with maps $\tilde{\psi} : C\to M$, and the correspondence is given by $\psi = \tilde{\psi}\circ \beta$.
Then $\psi(\im\alpha)=0$ if and only if $\alpha^*\psi = \psi\circ \alpha =0$ if and only if $\psi\in \ker\alpha^*$. Thus $\psi\in\ker\alpha^*$ if and only if
$\psi=\tilde{\psi}\circ \beta$ for some $\tilde{\psi}\in \operatorname{Hom}_R(C,M)$, i.e., $\psi \in \ker\alpha^*$ if and only if $\psi \in \im\beta^*$, as desired.
I believe I have settled this issue for myself, and the key ingredient is indeed the Dold-Kan correspondence. However, I find a synthesis of these two expositions to be much more comprehensible than anything nlab has to offer: (1), (2) They also both refer to the same book reference, namely this one, which is also helpful in understanding the result.
The basic idea is that any chain complex corresponds to a simplicial abelian group, and any simplicial abelian group $A$ has a corresponding chain complex $A_*$. The image of the degenerate simplices under this correspondence forms a subcomplex $DA_*$, and there is another subcomplex $NA_*$ which turns out to have the property that $NA_*\oplus DA_*\cong A_*$. In other words, $A_*/DA_*\cong NA_*$. The Dold-Kan correspondence theorem itself says that $A\to NA_*$ gives an equivalence of categories between chain complexes and simplicial abelian groups, which then gives a way to "normalize" certain chain complexes, including the chain complex defining the simplicial homology of a topological space.
In order to apply this to my situation, which involves cohomology rather than homology, it is necessary to "dualize" this result. The key insight, I think, is that actually any simplicial set $X$ can be "upgraded" into an abelian simplicial group $\mathbb Z (X)$, whose $n$-simplices $\mathbb Z(X)[n]$ are $\mathbb Z (X[n])$, which consists of formal sums $\sum_{x\in X}k_x x$ such that $k_x\in \mathbb Z$ and only finitely many of them are nonzero. Then any cochain complex $C^n(X,A)$ defined as $\lbrace f:X[n]\to A \rbrace$ can instead be thought of as $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$, and if $C_0^n(X,A)$ is the functions which are zero on any nondegenerate $n$-simplex, then $C^n(X,A)/C^0_n(X,A)\cong Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$ and the Dold-Kan correspondence gives a chain equivalence between $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$ and $ Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$, which makes the natural projection $C^n(X,A)\to C^n(X,A)/C^0_n(X,A)$ into a chain equivalence also.
After this, I found that it does appear necessary, or at least convenient, to go through $\mathcal E G$, the simplicial set with $\mathcal E G[n]=G^{n+1}$ and $d_i$, $s_i$ given by deletion and repetition of the $i$th element respectively, and whose geometric realization is $EG$. Even though this (or the bar resolution) may be viewed as a simplicial group, $G$ may not be abelian, so that isn't particularly helpful. It's better to think of it as a simplicial set and use a slightly modified version of the result above (essentially, the same formulation can be done for subcomplexes of $C^n(X,A)$ to "normalize" them). Then (in broad strokes) if $C^n(G,A)=\lbrace f:G^{n+1}\to A\text{ such that }gf=fg\text{ for all }g\in G \rbrace$, there is an injective map $C^n(G,A)/C^n_0(G,A)\to C^n_\text{cell}(EG,A)$, and an injective map $C^n_\text{cell}(BG,A)\to C^n_\text{cell}(EG,A)$; if $g$ acts trivially on $A$ then both of which have the same subcomplex as their image. The isomorphism follows from that, and if $G$ doesn't act trivially on $A$ then that first injective map can still be used to get an isomorphism to the cohomology with local coefficients instead.
Best Answer
In this case there is a standard 2-step resolution which may not be injective, precisely, but which suffices to compute the derived functor. You can read about it on the nlab, for example.
Note that we're talking about resolutions and injectivity in the category of inverse systems of abelian groups. So the resolution is really a chain complex of inverse systems -- maybe a bit more data than you were expecting.