Let $M$ be a manifold and let us consider an element $\Omega\in\Omega^{k}(M,L)$, namely a $k$-form that takes values on the flat line bundle $L\to M$. I want to know what this really means. Does it mean that for every small enough open set $U\subset M$ I can write $\omega$ as
$\Omega=\omega\otimes e$
where $e$ is any local section of $L$ based in $U$ and $\omega\in \Omega^{k}(M)$? Let $d_{\nabla}$ be the exterior covariant derivative acting on $\Omega^{k}(M,L)$. If $L$ is trivial (but with non-trivial connection $\nabla$) then there is an isomorphism $\Omega^{k}(M,L)\simeq \Omega^{k}(M)$. In this case we have that $d_{\nabla}\Omega = 0 \Leftrightarrow d\omega = 0$?, where $\omega$ is the image of $\Omega$ under this isomorphism?
Thanks.
As you said, $k$-forms with values in $L$ are locally $\omega\otimes e$, and since $L$ is at least locally trivial, they're locally just ordinary $k$-forms. The only difference is that we glue local forms together potentially with a twist coming from the transition maps of $L$.
For $L$ trivial, then, $\Omega^k(M,L)$ is not only isomorphic, but equal to $\Omega^k(M)$. The definition of smooth $k$-form doesn't use the connection. But an isomorphism such as you're interested in, in general nontrivial, is defined by sending any $d$-flat section to a $\Delta$-flat one, and extending by fiberwise linearity.