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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 $H^1(X; mathbb{Z}/kmathbb{Z})$ 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?

It's somewhat delicate here to get all the details right. First, $G$-bundles for a finite group $G$ are not required to be connected, so the relevant version of the classification of covering spaces is the disconnected version, which goes like this: the category of covering spaces of a nice connected space $X$ with basepoint $x$ is equivalent to the category of $pi_1(X, x)$-sets.

More explicitly, $n$-sheeted covers (possibly disconnected) are equivalent to actions of $pi_1(X, x)$ on $n$-element sets, or even more explicitly to conjugacy classes of homomorphisms $pi_1(X, x) to S_n$. Said another way, $n$-sheeted covers, possibly disconnected, are classified by the nonabelian cohomology set

$$H^1(X, S_n).$$

Among these, the connected covers correspond to the transitive actions, which are classified by conjugacy classes of subgroups of $pi_1(X, x)$ of index $n$. Among these, the regular covers correspond to normal subgroups.

Now, for $G$ a finite group, a $G$-bundle is more data than a $|G|$-sheeted cover: the fibers are equipped with a free and transitive right action of $G$ and everything has to be compatible with this. Said another way, $G$-bundles are equivalent to actions of $pi_1(X, x)$ on $G$ regarded as a right $G$-set, or more explicitly to conjugacy classes of homomorphisms $pi_1(X, x) to G$ (thinking of $G$ as a subgroup of $S_{|G|}$ to make the connection back to covers).

Given a finite regular $n$-sheeted cover $Y to X$ with corresponding normal subgroup $H = pi_1(Y, y)$ of $pi_1(X, x)$, we can think of this cover as a $G = pi_1(X, x)/H$-bundle, but not all $G$-bundles arise in this way (the monodromy map $pi_1(X, x) to G$ is not required to be surjective in general), and we only know that $G$ is some finite group of order $n$. Moreover, the data of a $G$-bundle includes the data of an isomorphism between $G$ and this quotient; it's not enough just to know that it exists.

So we can find a finite regular $n$-cover which is not a $mathbb{Z}/nmathbb{Z}$-bundle, even up to isomorphism of covers, by finding a group $pi_1(X, x)$ with a normal subgroup $H$ of index $n$ such that the quotient is not $mathbb{Z}/nmathbb{Z}$. A simple example is $X = T^2, pi_1(X, x) cong mathbb{Z}^2$; take $H = 2 mathbb{Z}^2$, so that the quotient is $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$.

And we can find a $mathbb{Z}/nmathbb{Z}$-bundle which is not a finite regular $n$-cover in the usual sense, again even up to isomorphism of covers, by finding a disconnected such bundle; for example, $X times mathbb{Z}/nmathbb{Z}$ for $n ge 2$ and any $X$.

Regular cyclic $q$-coverings are classified by normal subgroups $$Nhookrightarrow pi_1X to pi_1X/Ncong mathbb Z/q,$$

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.