Let $E$ be an elliptic curve over a field, $z \in E$ is a point, and $d \geq 1$. I consider a skyscraper sheaf $\mathcal{O}_z/m_z^d$, the evaluation map
$$ \operatorname{Hom}(\mathcal{O}, \mathcal{O}_z/m_z^d) \otimes \mathcal{O} \to \mathcal{O}_z/m_z^d $$
is surjective. Is it true that the kernel of this map is a vector bundle? In other words I have short exact sequence
$$ 0 \to P \to \operatorname{Hom}(\mathcal{O}, \mathcal{O}_z/m_z^d) \otimes \mathcal{O} \to \mathcal{O}_z/m_z^d \to 0, $$ where $P$ is a vector bundle of rank $d$.
Upd Matt's comment below make me think that the following could true. For any irreducible closed sub-scheme $D$ of codimension one we have short exact sequence $$ 0 \to \mathcal{O}(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0, $$ let me fix some $n>0$ then short exact sequence arising from evaluation map for $\mathcal{O}_{nD}$ is $$ 0 \to \bigoplus_{i=1}^n \mathcal{O}(-iD) \to \operatorname{Hom}(\mathcal{O}, \mathcal{O}_{nD}) \otimes \mathcal{O}_X \to \mathcal{O}_{nD} \to 0. $$
For any sheaf of $\mathcal O$-modules $\mathcal M$, the sheaf-Hom $Hom_{\mathcal O}(\mathcal O, \mathcal M)$ is naturally isomorphic to $\mathcal M$ (given by evaluation at the global section $1$ of $\mathcal O$).
The tensor product $\mathcal M \otimes_{\mathcal O} \mathcal O$ is also naturally isomorphic to $\mathcal M$ (the isomorphism being given by the $\mathcal O$-module structure on $\mathcal M$).
Putting these two together, we see that the evaluation map $Hom(\mathcal O,\mathcal M) \otimes \mathcal O \to \mathcal M$ is an isomorphism.
In particular, the kernel of your evaluation map is trivial. Thus it is a vector bundle, although possibly not the one you had in mind.
Added in response to comment below:
The above answer deals with sheaf Hom. This answer deals with global Hom, which is what the OP is interested in.
The tensor product $\mathrm{Hom}_{\mathcal O}(\mathcal O,\mathcal M) \otimes_k \mathcal O$ is a free $\mathcal O$-module, since it is just the tensor product of a $k$-vector space with $\mathcal O$. If $\mathcal M$ is coherent and we are on a projective variety, then $\mathrm{Hom}_{\mathcal O}(\mathcal O,\mathcal M)$ (which is just another way of writing the global sections of $\mathcal M$) is finite-dimensional, and so the tensor product is in fact free of finite rank.
The kernel of the evaluation map from the tensor product to $\mathcal M$ is thus a torsion-free $\mathcal O$-module. On a smooth curve, torsion-free coherent sheaves are locally free. (Because over a PID f.g. torsion-free modules are free.)
Thus the kernel in question is locally free.