Let $X$ be a smooth irreducible smooth projective curve. Let $L$ be a line bundle over X of degree $0$ such that $L^2\neq \mathcal{O}_X$.
Let $V\in Ext(L,L^{-1})$ and $W\in Ext(L^{-1},L)$. i.e.,
$0\rightarrow L\rightarrow V\rightarrow L^{-1}\rightarrow 0$ -----(1)
$0\rightarrow L^{-1}\rightarrow V\rightarrow L\rightarrow 0$------(2)
Consider $E=V\bigoplus W$.
How to prove that $End(E)$ has four generators, which in terms of block matrices can be described as $x= \left( \begin{array}{ccc} 0 & 0 \\ \gamma _2 & 0\end{array} \right), w_1= \left( \begin{array}{ccc} 0 & \gamma _1 \\ 0 & 0\end{array} \right), u= \left( \begin{array}{ccc} I & 0 \\ 0 & 0\end{array} \right), v= \left( \begin{array}{ccc} 0 & 0 \\ 0 & I\end{array} \right)$ where I is $2\times 2$ identity matrix, and $\gamma _1 $ and $\gamma_2$ coming from the identification of the line bundles in the exact sequence (1) and (2).
I have no idea how to prove this and what is $\gamma_1$ and $\gamma_2$.
First, some additional assumptions are necessary: we must assume that $V,W$ are not the split extensions $L \oplus L^{-1}$. Otherwise the statement isn't true (for example, take $V = W$ to both be split; then $\mathrm{End}(E)$ has 8 generators, not 4.)
First, note that the condition that $L$ is degree zero but $L^2 \ne \mathcal{O}_X$ implies that $$\mathrm{Hom}(L,L^{-1}) = \mathrm{Hom}(L^{-1},L) = 0.$$ On the other hand, since $X$ is projective, we have $$\mathrm{Hom}(L,L) = \mathrm{Hom}(L^{-1},L^{-1}) = k.$$ (In particular, the only morphisms are multiplication by a constant.) I will use these facts throughout.
Next, a useful Homological Lemma: Let
$$\begin{array}{ccccccccc} 0 & \to & A & \to & B & \to & C & \to & 0\\ & & \alpha \downarrow & & \beta\downarrow & & \gamma \downarrow & & \\ 0 & \to & A' & \to & B' & \to & C' & \to & 0 \end{array}$$
be a map of short exact sequences. The induced maps
$$0 \to \mathrm{im}(\alpha) \to \mathrm{im}(\beta) \to \mathrm{im}(\gamma) \to 0$$
are exact at the left and right, but not in the middle. Instead this is just a complex, and the homology in the middle is
$$H_{\mathrm{im}} = h\big(\mathrm{im}(\alpha) \to \mathrm{im}(\beta) \to \mathrm{im}(\gamma)\big) \cong \mathrm{im}(\delta),$$
where $\delta$ is the boundary map (coming from the snake lemma).
Proof omitted (but fun!). (Show that the boundary map factors as $\ker(\gamma) \twoheadrightarrow H_{\mathrm{im}} \hookrightarrow \mathrm{coker}(\alpha)$.)
Step 1. We show $\mathrm{End}(V) = k$, given by multiplication by a constant.
To see this, let $\phi : V \to V$ be any endomorphism. The composite map $L \hookrightarrow V \xrightarrow{\phi} V \twoheadrightarrow L^{-1}$ must be zero, which says that we get an induced map of short exact sequences
$$\begin{array}{ccccccccc} 0 & \to & L & \to & V & \to & L^{-1} & \to & 0\\ & & \downarrow & & \downarrow \phi & & \downarrow & & \\ 0 & \to & L & \to & V & \to & L^{-1} & \to & 0 \end{array}$$
Now, observe that the two unlabeled vertical arrows must both be multiplication by some constants $a,b \in k$. In particular, they are either isomorphisms, or zero. By subtracting $b \cdot \mathrm{id}$ throughout, we henceforth assume that the map $L^{-1} \to L^{-1}$ is zero (i.e. $b=0$); in particular the image of $\phi$ lands in $L$. (Our goal is now to show that $\phi = 0$.)
If $a \ne 0$, we can then rescale $\phi$ so that $L \to V \to L$ is the identity map -- but this would give a splitting of $V$. (This is why we need to assume $V$ is not split). So we must have $a = 0$. Now note that since $a=b=0$, the boundary map has the form $\delta : L^{-1} \to L$, hence must be zero, hence by the Homological Lemma, the sequence of images $$0 \to 0 \to \mathrm{im}(\phi) \to 0 \to 0$$ must be exact. But then $\mathrm{im}(\phi) = 0$. Done!
By symmetry, we have also shown $\mathrm{End}(W) = k$.
Step 2. We show that $\mathrm{Hom}(V,W) \cong k$. We have defining sequences
$$0 \to L \to V \to L^{-1} \to 0,$$ $$0 \to L^{-1} \to W \to L \to 0,$$
which gives a composite map $\psi : V \twoheadrightarrow L^{-1} \hookrightarrow W$, with kernel $L$ and cokernel $L^{-1}$. This is our generator.
Now let $\phi : V \to W$ be any morphism. Proceeding similarly to Step 1, observe that the composite map $L \hookrightarrow V \xrightarrow{\phi} W \twoheadrightarrow L$ must be zero, or else (after rescaling) we obtain splittings of $V$ and $W$. (In particular, $\phi$ has rank at most 1 at each point.) So we get a map of short exact sequences as in Step 1. But this time, the left and right vertical maps are already zero, which tells us that $\mathrm{im}(\phi) = \mathrm{im}(\delta)$, where $\delta : L^{-1} \to L^{-1}$ is multiplication by a constant.
Finally, note that this constant cannot be zero for $\psi$, since the kernel of $\psi$ is exactly $L$. So, to finish, observe that $\mathrm{im}(\phi - t \cdot \psi) = \mathrm{im}(\delta_\phi - t \cdot \delta_\psi)$, so by choosing $t$ appropriately, this is zero, hence $\phi = t \cdot \psi$.