Let $X$ be a finite CW complex. Swan's theorem provide an equivalence
$$
{\rm Vec}(X)\xrightarrow\sim{\rm ProjMod}(\mathop{\rm hom}\nolimits\_{\rm Top}(X,\mathbb{R}))
$$
between the category of **finite dimensional** vector bundles over $X$ and the category of finitely generated projective modules over the ring of continuous functions from $X$ to the reals. This isomorphism behaves well with the monoidal structure $\oplus$.


There is an intermediate step in this construction: The category ${\rm Vec}(X)$ of finite dimensional vector bundles over $X$ is equivalent to locally free modules of finite rank over the sheaf $C\_X(-)=\mathop{\rm hom}\nolimits\_{\rm Top}(-,\mathbb{R})$ on $X$.


The category ${\rm Cov}(X)$ of covers of $X$ is equivalent to the category of locally constant sheaves **of sets** on $X$. Is it possible to formulate this analogously to the above correspondence? So maybe locally constant sheaves are somehow special modules over $C\_X(-)$ and this category possibly corresponds to some special modules over $C\_X(X)$. Maybe this is also compatible with disjoint unions of coverings and sums of the corresponding modules. Maybe it is also necessary to require that the covering is **regular**.


(The **bold** things are edits made, partially based on the answers below.)



Swan-like theorem and covering spaces
If I can take only the finite covers, then yes, I think. (After all, Swan's theorem is a characterization of finite-dimensional vector bundles, not all vector bundles.) This is easier to do over $\mathbb{C}$ than over $\mathbb{R}$. In addition to the entire sheaf $C\_X(-)$, let $C(X) = C\_X(X)$ be the algebra of global continuous functions.


Since the module $M$ is locally free, what you want to do is to choose a basis for $C(U) = C\_X(U)$ for enough open sets $U$, and such that the bases agree when you restrict to smaller open sets. You could just ask for this directly, but there is an indirect algebraic condition that comes to the same thing. Namely, you can ask for $M$ to not only be finitely generated and projective, but also a semisimple commutative algebra over $C(X)$. This gives you the unordered basis in each fiber.


Over $\mathbb{R}$, it's not quite enough to require that $M$ be a semisimple algebra, because you could end up creating $C(Y,\mathbb{C})$ for a finite cover $Y$ of $X$. So, you could also impose the condition that $f^2 + g^2 = 0$ has no non-trivial solutions in $M$.


