Since you assumed that $X$ is smooth (less would be enough, but you need at least $S_2$), $G=j_*F_U$ is *reflexive*, that is $G^{\ast\ast}=G$ where $G^{*}=\mathscr Hom_X(G,\mathscr O_X)$ is the dual.

If two reflexive sheaves agree on an open set with a codimension $2$ complement, then they agree, so your question is equivalent to asking that $G$ be locally free.

If the rank of $F$ is $1$, then $G$ corresponds to a divisor and again since $X$ is smooth it is Cartier and hence $G$ is locally free.

If the rank of $F$ is at least $2$, then the locus where a reflexive sheaf is not locally free is at least $3$-codimensional (see Lemma 1.1.10 in Vector Bundles on Complex Projective Spaces, by Okonek, Schneider, Spindler).

So what you want can be done on curves and surfaces, but not necessarily on varieties of dimension $\geq 3$. An alternative way to get $G$ is to take the double dual of $F$, that is, $G=F^{\ast\ast}$. In other words, your question is whether that is locally free.

To complete the picture look at Example 1.1.13 in ibid. That shows that reflexive sheaves of rank $\geq 2$ are not always locally free.

To summarize:

- If $k=1$ or $\dim X\leq 2$ then the sheaf you are looking for is $j_*F_U$
- Otherwise, if $j_*F_U$ is locally free you've got it, if it is not, then there is no such sheaf.