In general, a twisted coefficient system on a manifold $M$ (also called a local system by more algero-geometrically minded people) is given by a representation of $\pi_1(M)$
(the holonomy representation, also called the monodromy representation by more algebrao-geometrically minded people). Conversely, any representation of $\pi_1(M)$ gives a twisted coefficient system on $M$.
(There is no need for $M$ to be a manifold here; any space for which the usual theory of $\pi_1$ and covering spaces goes through would be fine.)
If $V$ is the representation of $\pi_1(M)$, giving rise to the twisted coefficient system $\mathcal L$, then
there will be map $H^i(\pi_1(M), V) \to H^i(M,\mathcal L)$.
However, these will not be isomorphisms in general unless $M$ is aspherical,
i.e. if its universal cover $\tilde{M}$ is contractible, i.e. if $M$ is a $K(\pi,1)$ (for $\pi = \pi_1(M)$). (Here I am recalling the basic topological interpretation of group cohomology, which you can find in many places.)
What happens in general is that there is a spectral sequence (a special case of the Hochschild--Serre spectral sequence)
$$H^i(\pi_1(M), H^j(\tilde{M},V) ) \implies H^{i+j}(M,\mathcal L).$$
Note that if $M$ is hyperbolic (or, more generally, negatively curved), then
its universal cover is contractible, and so one does get your desired isomorphism.
Added: To get a feeling for what can happen if $\tilde{M}$ is not contractible, you can consider the case when $M = \mathbb RP^2$, so that
$\tilde{M}$ is $S^2$ and $\pi_1(M)$ is cyclic of order $2$, acting on $S^2$ by the antipodal map. Take $V$ to be the trivial representation (over $\mathbb Z$, or $\mathbb Z/2\mathbb Z$, say). Then the preceding spectral sequence gives a way to compute the cohomology (with $\mathbb Z$-coefficients, or $\mathbb Z/2\mathbb Z$-coefficients) in terms of the group cohomology of the cyclic group of
order two acting on the cohomology of the sphere. (So it acts trivially
on $H^0$, and by $-1$ on $H^2$.)
Of course, you could do the analogous computation for $\mathbb R P^3$ as well
(which is more directly relevant to your question).
Let $P$ be a principal $G$-bundle, $\rho:G\to GL(V)$ a finite dimensional representation of $G$, $E = P \times_G V$ the associated vector bundle. To any principal connection $\Phi$ on $P$ is associated an induced linear connection $\bar \Phi$ on $E$. Conversely, any linear connection on a vector bundle $E$ is induced from a unique principal connection on the linear frame bundle $GL(\mathbb R^n, E)$ of $E$.
You can find all the details here: Topics in Differential Geometry - P. W. Michor.
Best Answer
Your general statement is simply false. For instance, if the action is trivial, your statement would say that the homology of any space agrees with the homology of its universal cover, which is ridiculous.
What is true is that if $A = \Bbb Z[\pi_1(X)/H]$ as a $\pi_1(X)$-module, and $p: X' \to X$ is the covering space associated to the subgroup $H \subset \pi_1(X)$, then $\mathcal A = \tilde X \times_{\pi_1 X} A$ gives a local coefficient system over $X$ (if you like to think of a local coefficient system as being a bundle of abelian groups over $X$). Then we have $$H_*(X;\mathcal A) \cong H_*(X'; \Bbb Z).$$
What this says is that a very specific local coefficient system gives the homology of this covering space. Not that you can compute the homology of general coefficient systems by passing to a covering space.