Let's say that $X$ is an integral scheme of finite type over a field and $Y \subset X$ is a closed subscheme. Given a vector bundle $E$ on $Y$, is $E$ the restriction to $Y$ of a vector bundle on a neighborhood $U$ of $Y$ in $X$? 



Extending a vector bundle to a torsion free sheaf
No, this is not even true for line bundles.


For an example, let $X = \mathbb{P}^2$ and $Y$ a smooth curve in $X$ of genus $>0$ (over an algebraically closed field). Since $X$ is smooth, any line bundle on an open set $U$ extends to a line bundle on $X$ so the map $\operatorname{Pic}(X) \to \operatorname{Pic}(U)$ is surjective. Since $\operatorname{Pic}(X) \cong \mathbb{Z}$, it follows that the image of $\operatorname{Pic}(U)$ in $\operatorname{Pic}(Y)$ is of rank $1$ and is independent of $U \supset Y$. Since $\operatorname{Pic}(Y)$ is not even finitely generated we see that there exist (many) line bundles on $Y$ which do not extend to any open $U \supset Y$.


