Classifying Covering Spaces using First Cohomology I am familiar with the classification of covering spaces of a space $X$ in terms of subgroups of $\pi_1(X)$ (up to conjugation). However, if $X$ is a manifold, I know that $H^1(X; G)$ classifies G-bundles over $X$ (using Cech cohomology here). I think finite regular covering spaces are $\mathbb{Z}/k \mathbb{Z}$-bundles; regular means that the deck transformations act transitively on the fiber (and regular covers correspond to normal subgroups of $\pi_1(X)$). 
Does this mean that $H^1(X; \mathbb{Z}/k\mathbb{Z})$ is in bijection with k-sheeted regular covering spaces over $X$. I could not find such a statement anywhere and so am a bit suspicious. 
Also, if this is correct, what does $H^1(X; \mathbb{Z})$ classify? I'm not sure what a $\mathbb{Z}$-bundle is - what has automorphism group equal to $\mathbb{Z}$? 
Also, $H^1(X; \mathbb{Z}) = [X, S^1]$ so if $H^1(X; \mathbb{Z})$ classifies some kind of bundles, there should be universal bundle over $S^1$ which pulls back to these bundles. What is this bundle? 

Regular cyclic $q$-coverings are classified by normal subgroups $$N\hookrightarrow \pi_1X \to \pi_1X/N\cong \mathbb Z/q,$$
hence there is a natural isomorphism to $Hom(\pi_1X,\mathbb Z/q)$ mod automorphism of $\mathbb Z/q$.
Note that $K_q=K(\mathbb Z/q,1)$ is universal in the sense that the map $$[X,K_q] \to Hom(\pi_1X,\mathbb Z/q) = H^1(X,\mathbb Z/q),$$ $$ [f]\mapsto f_*$$ is an isomorphism. In particular you see that every cyclic $q$-covering is obtained by pulling back the universal cover of $K_q$.
Corollary If $X$ is a smooth orientable manifold, then all infinite cyclic covers (or cyclic which factor through $\mathbb Z$) are obtained by cutting and gluing along a hypersurface.
