"Local free"ness in vector bundles and projective modules Swan's Theorem tells us that (real) vector bundles on $X$ are the same as finitely-generated projective modules over $C(X)$ (continuous $\mathbb{R}$-valued functions on $X$. And vector bundles are thought of as locally free sheaves.
There's also a local freeness condition for modules over any commutative ring $R$. An $R$-module $P$ is locally free if the localization $P_\mathfrak{m} := R_\mathfrak{m} \otimes_R P$ is free as a $R_\mathfrak{m}$-module for every $\mathfrak{m} \in \text{MaxSpec}(R)$.
Theorem: If $P$ is a finitely-presented $R$-module, then $P$ is locally free iff $P$ is projective.
So "locally free" sheaf is the same as "locally free" module (except you need some kind of finite presentation condition). What does this have to do with trivializations? Does a trivialization of a vector bundle correspond to a localization of its module of global sections?
More specifically: a trivialization of a bundle $E$ is an open set $U$ over which the bundle is a Cartesian product $U\times \mathbb{R}^k$. Does this somehow correspond to a maximal ideal $\mathfrak{m}$ of $C(X)$ so that $\Gamma(E)_\mathfrak{m}$ is a free module?
 A: A locally free sheaf is only the same as a locally free module over an affine scheme/variety.
There's no finite presentation condition required.
A trivialization of a vector bundle is a cover of your space by Zariski open sets such that the restriction of your bundle to each open set in the cover is isomorphic to the trivial vector bundle (of the same rank).
The connection with locally free sheaves, is that any vector bundle is determined by its sheaf of sections, and vice versa. For a trivial vector bundle $U\times K^n$ over $U$ ($K$ is a field), a section $U\rightarrow U\times K^n$ is the same as giving an $n$-tuple of functions $(f_1,\ldots,f_n)\in\mathcal{O}_U(U)^n$, so for trivial vector bundles, its sheaf of sections is just the free sheaf of rank $n$. Hence, any trivialization of a vector bundle corresponds to a trivialization of its sheaf of sections (ie, an open cover where on each open set the sheaf of sections is actually free). In terms of commutative algebra, if the base space was affine, and you choose a trivialization by affine open sets, then yes the restriction of your sheaf of sections to each affine open in the trivialization, corresponding to a localization of the corresponding locally free module, will give you actually a free sheaf, corresponding to a free module.
