Vector Bundle Locally Free Sheaf Let $f: X \to Y$ a morphism of ringed spaces and $E$ a vector bundle over $Y$ for finite rank $r$.
My questions are:


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*Why $E$ can be interpreted as locally free sheaf of rank $r$, therefore for each $y \in Y$ there exist an open set $U$ that contains $y$ such that $E \vert _U \cong \mathcal{O}_Y ^{\otimes r}\vert _U$

*Conversly, how can a locally free sheaf of rank $r$ turned to a vector bundle formally correct?
Background of my question: How to interpret $f^*E$? Problem:$f^*$ accepts as arguments only sheafs on $Y$...
 A: The idea is to interpret the sheaf of section of the bundle $E$ as the locally free sheaf you mean. Since $E$ is a vector bundle, over small enough $U \subset Y$, $E|_U \simeq U \times k^n$, and the sections of the bundle $U \times k^n \to U$ correspond exactly to maps $U \to k^n$, whence we get isomorphism with $\mathcal{O}_U^n$.
Turning locally free sheaf $\mathcal{E}$ into a vector bundle $E \to Y$ is similarly simple: if our sheaf is free over each of $U_i$, and $\bigcup U_i = Y$, then we just need to glue $U_i \times k^n \to U_i$ over various $U_i$. Therefore, we need a family of gluing isomorphisms $u_{ij}: (U_i \cap U_j) \times k^n \to (U_j \cap U_i) \times k^n$. 
Here's how we get them: we have local isomorphism $f_i: \mathcal{E}_{U_i} \to \mathcal{O}_{U_i}^n$ from the definition of "locally free". Restricting to intersection $U_i \cap U_j$, we get $f_{ij} = f_j|_{U_i \cap U_j} \circ f_i|_{U_i \cap U_j}^{-1}: \mathcal{O}_{U_i}^n|_{U_i \cap U_j} \to \mathcal{O}_{U_j}^n|_{U_i \cap U_j}$. Every such map is induced by a map $u_{ij}: (U_i \cap U_j) \times k^n \to (U_j \cap U_i) \times k^n$ (why?), and this is our gluing isomorphism.
A: Consider an arbitrary (topological) $k$-vector bundle $\pi: E \to X$. For each $U \subset X$, we'll assume that $\mathcal{F}_X(U)$ be the set of continuous maps $U \to k$. It isn't hard to see that $\mathcal{F}_X(U)$ is a sheaf of rings, for we can take sums, products, and restrictions of functions in the usual way. 
We'll assume that $\mathcal{O}_X$ is a subsheaf of $\mathcal{F}_X$; otherwise it is not always possible to construct a locally free sheaf correponding to the vector bundle $\pi$. 
Now if we let $\mathcal{E}(U)$ be the set of continuous sections $f: U \to \pi^{-1}(U)$ (where "section" means that $f(u)$ lies "above" $u$ in the sense that $\pi f(u) = u$) then I'll show $\mathcal E$ is an $\mathcal{O}_X$-module that's locally free of rank $\dim E$. 
To see that it's an $\mathcal{O}_X$-module note that for $\psi \in \mathcal E(U)$, we have $$\psi(u) = (u,v) \in \{u\} \times k^n.$$ For $f \in \mathcal{O}_X(U)$, since $f(u) \in k$ we can use scalar multiplication in $k^n$ to define $$f\psi(u) = (u, f(u)\cdot v) \in \{u\} \times k^n.$$ Since $E$ is locally a cartesian product, $\psi$ is locally a cartesian product of continuous coordinate functions. Since multiplication of continuous functions is continuous, it follows that $f \psi \in \mathcal E(U)$, making $\mathcal E$ an $\mathcal O_X$-module.
Similarly, to check that the sheaf we have constructed is locally free, we'll use the fact that $E$ is locally homeomorphic to a product space. A continuous section $U \to U \times k^n$ is the same thing as $n$ continuous maps $U \to k$, hence $\mathcal{E}|_U \simeq \mathcal{O}_U^n$ as an $\mathcal{O}_U$-module.
The (possibly) surprising fact is that the converse is true: a locally-free $\mathcal{O}_X$-module of rank $r$ corresponds uniquely to a vector bundle $E$. To reiterate, all of this only makes sense if we can interpret $\mathcal{O}_X$ as a sheaf of continuous $k$-valued functions.
