Counterexample: Homology with Local Coefficients versus Homology with Module Coefficients Definitions Let $X$ be a nice space with universal covering $\widetilde{X}$.


*

*Homology $H_*(X,R)$ with Ring Coefficients $R$ is the homology of
\begin{align*}
C_n(X;R) :=&\; \text{ free left $R$-module with basis $S_n(X)$}, \\
\cong&\; R \otimes_{\mathbb{Z}} C_n(X)
\end{align*}
where $C_*(X)$ is the singular chain complex.

*Homology $H_*(X;M)$ with Module Coefficients $M$, where $M$ is a left $R$-module, is the homology of
$$
C_n(X;M) := M \otimes_R C_n(X;R)
$$

*Homology $H_*(X;A)$ with Local Coefficients in $A$, where $A$ is a  left $R[\pi_1X]$-module, is the homology of
$$
C_n(\widetilde{X};R) \otimes_{R[\pi_1X]} A
$$
where we view $C_n(\widetilde{X};R)$ as a right $R[\pi_1X]$-module using the monodromy action on $\widetilde{X}$.
Question: Is the local homology with local coefficients given by a left $R[\pi_1X]$-module $A$ (def. 3) the same as the homology with module coefficients in $A$, viewed as a $R[\pi_X]$-module (def. 2)? In other words, is the expression
$$
H_n(X;A)
$$
even well-defined? If not, what is a counter example?
 A: They are distinguished by their $\pi_1X$ action. Your definition of coefficients in a module, is the so-called untwisted homology by:
$$
C_n(X;M) := M \otimes_R C_n(X;R) = M \otimes_R R^{|n|} = M^{|n|},
$$
where $|n|$ denotes the number of $n$-cells of $X$ with no $\pi_1$ action on it. (3) in contrast is the twisted homology. As suggested by the name, (2) is the "regular" case, and (3) generalizes it by a  non-trivial $\pi_1X$ action.
So if you want, define $M$ as a $R[\pi_1X]$-module by the trivial action, which then gives module homology as a special case of local coefficients. You can read about this in Hatcher's Algebraic Topology, more specifically you are going to learn if the action on $A$ is given by $a:\pi_1 X\to Aut(A)$, it suffices to look at a cover corresponding to $ker(a)$ while tensoring over $R[\pi_1(X)/ker(a)]$ the group ring over the deck transformations. In the trivial action case, that gives you the equivalence (up to the left- and right-stuff which you have to be careful with).
As you noticed correctly, this potentially leads to confusion. When being sloppy we get weird looking things like
$$
H_1(S^1;\mathbb Z/3) = \mathbb Z/3 \neq 0 = H_1(S^1;\mathbb Z/3),
$$
as computed here Homology with twisted coefficients of the circle . The difference is that on the left hand side I interpreted $\mathbb Z/3$ as a $\mathbb Z$-module, while on the right hand side I mean a module with a non-trivial fundamental group action, that is a $\mathbb Z[\mathbb Z]$-module with non-trivial $\mathbb Z$-action. 
Therefore it is common practice to potentially think of every homology coefficients $A$ as $\mathbb [\pi_1X]$-modules via $a: \pi_1X \to Aut(A)$ and write 
$$
H_i(X;A),
$$
if $a=1$ and otherwise specify
$$H_i(X;a),$$
by explicitely mentioning the twisting. However if the twisting is clear from context, often the former is (ab)used anyway.
