= Solution
For the <free presentation> $G=F/R$, conjugation in $F$ gives the <relation module>
$$
R_{\mathrm{ab}}=R/[R,R],
\qquad g\cdot(r[R,R])=frf^{-1}[R,R],
$$
where $f\in F$ is any lift of $g\in G$. A different lift differs by an element of $R$, whose inner conjugation acts trivially on the abelianization, so this is a well-defined $\mathbb ZG$-module action.
Choose free generators $x_1,\ldots,x_n$ of $F$. The <presentation relation sequence> becomes
$$
0\longrightarrow R_{\mathrm{ab}}
\xrightarrow{j}(\mathbb ZG)^n
\xrightarrow{d_1}\mathbb ZG
\xrightarrow{\varepsilon}\mathbb Z
\longrightarrow0,
$$
where $d_1(e_i)=\bar x_i-1$. In <Fox calculus>, the first map is
$$
j(r[R,R])=
\sum_{i=1}^n\overline{\frac{\partial r}{\partial x_i}}e_i.
$$
The Fox identity $r-1=\sum_i(\partial r/\partial x_i)(x_i-1)$ gives $d_1j=0$, and the standard lifting argument in the free group proves exactness.
Split the sequence at the augmentation ideal $I_G$. Applying $\operatorname{Hom}_{\mathbb ZG}(-,M)$ to
$$
0\longrightarrow R_{\mathrm{ab}}\longrightarrow(\mathbb ZG)^n\longrightarrow I_G\longrightarrow0
$$
and using projectivity of $(\mathbb ZG)^n$ gives
$$
M^n\longrightarrow\operatorname{Hom}_G(R_{\mathrm{ab}},M)
\longrightarrow\operatorname{Ext}^1_{\mathbb ZG}(I_G,M)\longrightarrow0.
$$
The other short exact sequence $0\to I_G\to\mathbb ZG\to\mathbb Z\to0$ identifies the last group with $H^2(G,M)$. A <one-cocycle> on $F$ is a derivation and is determined freely by its values on $x_1,\ldots,x_n$, so $H^1(F,M)$ is $M^n$ modulo principal derivations. Principal derivations vanish on $R$, and restriction sends a derivation $d$ to the $G$-map $r[R,R]\mapsto d(r)$. The preceding cokernel sequence therefore descends to the <Mac Lane exact sequence for a free presentation>
$$
H^1(F,M)\longrightarrow\operatorname{Hom}_G(R_{\mathrm{ab}},M)
\longrightarrow H^2(G,M)\longrightarrow0.
$$
The left map need not be injective. Take $F=\langle t\rangle\cong\mathbb Z$, $R=\langle t^m\rangle$, $G\cong C_m$, and the trivial module $M=\mathbb Z/m\mathbb Z$, where $m>1$. Then $H^1(F,M)\cong M$, while restriction sends the derivation determined by $d(t)=a$ to
$$
d(t^m)=ma=0.
$$
\b[Thus the left map is zero although its domain is nonzero, and $R$ is nontrivial.]
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