Filling a gap of an example I got the following problem:
Let $K$ be a field. If we consider the ring $$R=\begin{pmatrix} K & 0\\ K^{(\mathbb{N})} & K \end{pmatrix}$$ and the idempotents $\epsilon_{a}=\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}$ and $\epsilon_{b}=\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$ and let $\epsilon_{1}$ the first entry at the canonical basis of $K^{(\mathbb{N})}$, then put $S=R\begin{pmatrix} 0 & 0\\ \epsilon_{1} & 0 \end{pmatrix} =\begin{pmatrix} 0 & 0\\ K \epsilon_{1} & 0 \end{pmatrix}$. Set $M_{1}= R \epsilon_{a}$ and $M_{2}= \frac{R \epsilon_{a}}{S}$. What I want to prove is that $\mathrm{Hom}_{R}(M_{1}, M_{2})=K$ and $\mathrm{Hom}_{R}(M_{2},M_{1})= 0$.
For me this has maybe something related to $M_{1}$ and $M_{2}$ are simple. So far I have noted that $R \cong \epsilon_{b}$ and
$$R= \epsilon_{a} \oplus \epsilon_{b}= \begin{pmatrix} K & 0\\ K^{\mathbb{N}} & 0 \end{pmatrix} \oplus \begin{pmatrix} 0 & 0\\ 0 & K \end{pmatrix}.$$
For proving, $Hom_{R}(M_{1}, M_{2}) \cong K$ lets take $g \in Hom_{R}(M_{1},M_{2})$ and also consider the projection $\pi:M_{1} \to M_{2}=\frac{M_{1}}{S}$ so as we can factor this morphism $h$ into applying a morphism $f: M_{1} \to M_{1}$ and then $\pi$ but $f$ consist only in multiplying any $\begin{pmatrix} x & 0\\ \bar{v} & 0 \end{pmatrix} \in M_{1}$ with $\begin{pmatrix} a & 0\\ 0 & 0 \end{pmatrix}$ where $a \in K$. What I need to prove to show that $Hom_{R}(M_{1}, M_{2}) \cong K$ is that $a \in K$ in $f$ is unique for the factorization $h=\pi f$. 
We showed in the comments that $\operatorname{Hom}_R(M_2, M_1) \not= 0$. Namely the map $M_1 \rightarrow M_1$ defined by sending $A \in M_1$ to $\epsilon_aA \in M_1$ restricts to $0$ on $S \subseteq M_1$, so that map factors through $M_2 = M_1/S$ and is not zero on $M_2$ because it fixes $\epsilon_a \in M_1 \setminus S$.
For the second part of your question, you wondered if $\operatorname{Hom}_R(M_1, M_2) \cong K$.
This is also false. For example, an easy consideration comes from cardinality. For any $a \in K, b \in K^{(\mathbb{N})}$, we get a map sending an element $A \in M_1$ to the coset $\begin{bmatrix} a & 0 \\ b & 0\end{bmatrix}\cdot A + S \in M_2$. If $b, b'$ are elements that differ away from the first coordinate of $K^{(\mathbb{N})}$, then the corresponding maps are distinct, by definition of $S$. There are way to many of these homomorphisms to be in correspondence with elements of $K$. If $K$ is finite, there are already uncountably many of these homomorphisms.