I think I have a proof of:
If $R$ is noetherian, $M$ finitely generated and $N$ arbitrary then
$\mathrm{Ass}\;\mathrm{Hom}(M,N)=\mathrm{Supp}\;M\cap\mathrm{Ass}\;N$.
First: let's prove that $\mathrm{Ass}\;\mathrm{Hom}(M,N)\subset\mathrm{Supp}\;M$. Take $p\in\mathrm{Ass}\;\mathrm{Hom}(M,N)$, I want $\mathrm{Ann}\;M\subset p$. Take $x\in\mathrm{Ann}\;M$ and suppose $p=\mathrm{Ann}\;\varphi$. Then $x\varphi(m)=\varphi(xm)=\varphi(0)=0$ and so $x\in p$.
Second: $\mathrm{Ass}\;\mathrm{Hom}(M,N)\subset\mathrm{Ass}\;N$.
Suppose $p\in \mathrm{Ass}\;\mathrm{Hom}(M,N)$, this is equivalent to $PR_P\in\mathrm{Ass}\;\mathrm{Hom}(M_p,N_p)$ that is equivalent to $\mathrm{depth}\;\mathrm{Hom}(M_p,N_p)=0$.
In the same way $p\in\mathrm{Ass}\;N$ is equivalent to $\mathrm{depth}\;N_p=0$
So I want to prove: let $(R,m,k)$ local noetherian and $M\neq0$ finitely generated then $\mathrm{depth}\;\mathrm{Hom}(M,N)=0$ implies $\mathrm{depth}\;N=0$.
Suppose that $x\in m$ is a $N$-regular element. Take $\varphi\in\mathrm{Hom}(M,N)$, suppose $x\varphi=0$, then $x\varphi(m)=0$ for all $m$ and so $\varphi(m)=0$ for all $m$ because $x$ is $N$-regular and so $\varphi=0$, that implies $x$ is $\mathrm{Hom}(M,N)$-regular, contradiction.
Third: $\mathrm{Supp}\;M\cap\mathrm{Ass}\;N\subset\mathrm{Ass}\;\mathrm{Hom}(M,N)$.
As before I can suppose $(R,m,k)$ local noetherian, $M$ non zero finitely generated. It's enough to prove that if $m\in\mathrm{Ass}\;N$ then $m\in\mathrm{Ass}\;\mathrm{Hom}(M,N)$.
If $m\in\mathrm{Ass}\;N$ then I have an injection $k\rightarrow N$. Applying $\mathrm{Hom}(M,.)$ we obtain an injection $\mathrm{Hom}(M,k)\rightarrow\mathrm{Hom}(M,N)$. I want to prove that there is an injection $k\rightarrow\mathrm{Hom}(M,k)$. Suppose $M=_R$ is a minimal system of generators. Take $\bar{r}\in k$ and define $\varphi_{\bar{r}}:M\rightarrow k$ in the following way: $\varphi_{\bar{r}}(m_1)=\bar{r}$ and $\varphi_{\bar{r}}(m_i)=0$ if $i\neq1$. The function $\bar{r}\mapsto\varphi_{\bar{r}}$ is the required injection.
$\varphi_{\bar{r}}$ is well-defined: suppose $\sum_i r_im_i=0$, by Nakayama this implies $r_i\in m$. $\varphi_{\bar{r}}(\sum_i r_im_i)=r_1\bar{r}=0$ because $r_1\in m$.