How do we get this quotient $\textrm{Ext}^1(N,M)/\textrm{Hom}(N,M)$? If $0\longrightarrow M\longrightarrow E\longrightarrow N\longrightarrow 0$ is a short exact sequence of vector bundles on a surface.  Here $M$ and $N$ are line bundles, and so rank $ E$=2. Also, if $L$ is an ample line bundle, we have that $\mu_L(M)>\mu_L(E)>\mu_L(N)$.
My question is this : Why does $\textrm{Ext}^1(N,M)/\textrm{Hom}(N,M)$ make sense?
From the initial exact sequence, by applying the functor Hom$(N,\_)$, we get the sequence $0\longrightarrow \textrm{Hom}(N,M)\longrightarrow \textrm{Hom}(N,E)\longrightarrow \textrm{Hom}(N,N)\longrightarrow \textrm{Ext}^1(N,M)\longrightarrow..$.
Similarly in the case if we apply Hom$(\_,M)$, we get $0\longrightarrow \textrm{Hom}(N,M)\longrightarrow \textrm{Hom}(E,M)\longrightarrow \textrm{Hom}(M,M)\longrightarrow \textrm{Ext}^1(N,M)\longrightarrow..$.
So how do we get this quotient?

As stated in the comments, this does not answer the question at all, but rather one that arose there.
Suppose our category had enough projectives. Choose a projective resolution $d_{\bullet}\!: P_{\bullet} \to N$. Using projectivity of $P_0$, resp. projectivity of $P_1$ and $\text{im}(f_0 \circ d_0) \subseteq \ker(E \to N) = M$, we get lifts of the identity
$$\require{AMScd}
\begin{CD}
P_2 @>d_1>> P_1 @>d_0>> P_0 @>>> N @>>> 0 \\
@VVV @VVf_1V @VVf_0V @VV\text{id}V \\
0 @>>> M @>>> E @>>> N @>>> 0.
\end{CD}
$$
It follows that $E \cong M \amalg_{P_1} P_0$ is the pushout. On the other hand, consider the complex
$$0 \to \text{Hom}(P_0,M) \xrightarrow{d_0^*} \text{Hom}(P_1,M) \xrightarrow{d_1^*} \text{Hom}(P_2,M) \to \ldots$$
By construction, $f_1 \in \ker(d_1^*)$. This defines the class of our extension in $\text{Ext}^1(N,M) = H^1(\text{Hom}(P_{\bullet},M))$. Accordingly, it splits iff $f_1 \in \text{im}(d_0^*)$, and in fact, we have a bijection between $(d_0^*)^{-1}f_1$ and splittings of our extension*. Then $\text{Hom}(N,M) \cong \ker(d_0^*)$ acts on this simply transitively by addition.
*The splittings $M \amalg_{P_1} P_0 \cong E \to M$ are precisely given by $(\text{id}_M,\varphi)$ with $\varphi \circ d_0 = f_1$, that is, such that $\varphi \in (d_0^*)^{-1}f_1 \subseteq \text{Hom}(P_0,M)$.
