I indeed found this question Is a covering space of a manifold always a manifold. However I do not know the concepts here used. As far as I know I just need to present a suitable atlas for the covering space as I do not know what funcdamental groups, second countability etc are.
Suppose that we have a covering space $p:E\to B$ where $B$ is a manifold that admits a smooth structure. Then $p$ is a continuous surjection. Also, for all points $b \in B$ there exists as neighborhood $U$ such that $p^{-1}(U) = \sqcup _{\beta} V_{\beta} $ is the disjoint union of open sets in $E$. Now $B$ admits a smooth structure so we can find an atlas $\{ U_{\alpha},f_{\alpha} \}$ for $B$ where $U_{\alpha}$ are open subsets of $B$ homeomorphically mapped to $\mathbb R^n$, where $f_{\alpha _1} \circ f_{\alpha_2}^{-1}$ is a differentiable map.
Now consider the functions $$g_{\alpha} = f_{\alpha}\circ p:E\to \mathbb R^n.$$ These functions map the open sets $p^{-1}(U_{\alpha})=\bigsqcup \limits_{\beta} V^{\alpha}_{\beta} $ to $f_{\alpha}( U_{\alpha})$. We know that $p^{-1}(U_{\alpha})$ is open because $p$ is continuous. Also $f_{\alpha} (U_{\alpha})$ is an open set. Now I need to show that $g_{\alpha}$ are homeomorphsims and that the change of coordinate functions
$$g_{\alpha_1} \circ g_{\alpha_2}^{-1}$$
are differentiable and I am stuck here. It would be nice if someone can help me out to finish the argument. As I said before I am probably missing some of the big machinery that makes this question easier to answer.
As a sidenote, there is some other point that I am not sure about. I implicitly assumed that the open sets $U_{\alpha} \subset B$ are precisely those sets that are mapped to the disjoint union of open sets in $E$ by $p$, but this need not be the case. I thought maybe I could choose an atlas for $B$ so that all $U_{\alpha}$ are at least contained in such neightborhoods.
Best Answer
What you have done so far is (mostly) correct, and no heavy machinery is required to finish the proof. Your sidenote at the end is an important point, but it is easy to resolve: If $(U_\alpha,f_\alpha)$ is a chart about $b\in B$, and $W_\alpha$ is an evenly covered neighborhood of $b\in B$ (i.e., $p^{-1}(W_\alpha)$ is a disjoint union of open sets each homeomorphic to $W_\alpha$), then $U_\alpha \cap W_\alpha$ is also evenly covered (easy to show), and $(U_\alpha\cap W_\alpha, f_\alpha)$ is a chart about $b\in B$ (also easy to show). So just replace all of your $U_\alpha$ with $U_\alpha'=U_\alpha\cap W_\alpha$.
Now your pretty much done, you just have to define your charts on the cover a bit differently: $g_\alpha^\beta=f_\alpha\circ p\mid_{V_\alpha^\beta}:V_\alpha^\beta\to f_\alpha(p(V_\alpha^\beta))\subset \mathbb{R}^n$, i.e., the open set domain for your chart is just one of the $V_\alpha^\beta$, not the entire disjoint union. This is a homeomorphism because $U_\alpha'$ evenly covered means $p\mid_{V_\alpha^\beta}:V_\alpha^\beta\to U_\alpha'$ is a homeomorphism, and because $(U_\alpha',f_\alpha)$ is a chart.
Now you just have to show that transition maps are smooth. This is easy, if slightly tedious, so I'll let you work it out.