$\newcommand{\Z}{\mathbb{Z}}\newcommand{\T}{\mathbb{T}}\newcommand{\R}{\mathbb{R}}$Motivated by this MO question, I was trying to understand this nLab section about computing cohomology with local coefficients. To see if I got it right I wanted to compute a trivial example. But I quickly ran into trouble.
With notation following the nLab entry, let's take $X=\T^2$, be a two-torus and $A\to X$ be the trivial bundle with $\bar{A}:=\R$ fibres: $A\cong X\times\bar{A}=\T^2\times \R$. The connection on $A$ is the trivial connection, so that $\mathrm{d}_\nabla = \mathrm{d}:\Omega^p_A\to\Omega^{p+1}_A$ is just the de Rham differential. Therefore the cohomology we are computing is $$H^p(X;\mathscr{A}) \cong H^p(X;A,\nabla) \cong H^p(\T^2;\T^2\times\R,0) \cong H^p(\T^2;\R),$$ just the cohomology of the two-torus with real coefficients.
Now, following nLab, substituting my trivial example, we have that the universal covering space is $\widetilde{X}=\R^2$, $\pi:=\pi_1(X)=\Z^2$ and the deck transformations are just translations $\R^2\to\R^2$. The trivial bundle over the covering space is $\widetilde{A}\cong\widetilde{X}\times \bar{A} = \R^2\times\R$. The action of $\pi$ on $\Omega^p(\widetilde{X})$ is given by the pullback along a deck transformation, so, by the pullback along a translation. In this case, the $\pi$-invariant piece of $\Omega^p(\widetilde{X})$, $\Omega^p(\widetilde{X})^\pi$ is simply translation-invariant $p$-forms on $\R^2$, so just periodic $p$-forms. So, $\Omega^p(\widetilde{X})$ trivially splits into $\Omega^p(\widetilde{X})^\pi\oplus\Omega^p(\widetilde{X})^\hat{\pi}$, as periodic$\oplus$non-periodic $p$-forms, and hence so does $\Omega^p_{\widetilde{A}}(\widetilde{X}):=\Omega^p(\widetilde{X})\otimes\bar{A}$. This means that the conditions for Proposition 4.2. are met and hence gives that $$H^p(X;\mathscr{A})\cong \left(H^p(\widetilde{X})\otimes\bar{A}\right)^\pi.$$
In this specific case this implies: $$ H^p(\T^2;\R) \cong H^p(X;\mathscr{A})\cong \left(H^p(\widetilde{X})\otimes\bar{A}\right)^\pi \cong \left(1\otimes\R\right)^\pi \cong 1.$$ This is, of course, obviously wrong! The question is, where did I mess up? How would things change if I actually had a non-trivial bundle to begin with?