Action on group $\operatorname{Ext}^i(\mathcal{L}, \mathcal{M})$ by scalar multiplication Let $X$ be a proper scheme over field $k$ and $\mathcal{L}, \mathcal{M}$ two invertible $\mathcal{O}_X$-modules. Then $Hom_{\mathcal{O}_X}(\mathcal{L}, \mathcal{M}) \cong Hom_{\mathcal{O}_X}(\mathcal{O}_X, \mathcal{M}\otimes  \mathcal{L}^{\vee}) \cong H^0(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$.
Therefore derived functors coinside as well as we assumed $X$ sufficiently nice:
$\operatorname{Ext}^i(\mathcal{L}, \mathcal{M}) \cong H^i(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$.
The right hand side has $ H^i(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$ a natural structure of a $k$ vector space, therefore we can talk about subspaces, multiplication by scalars form $k$ and the whole other basic linear algebra stuff.
On the other hand the Abelian group $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ has an interpretation as set of all extension classes
$$0 \to \mathcal{L} \to ? \to \mathcal{M} \to 0$$
where two classes are considered in  $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ as equal if there exist commutative diagram between the two exact sequences such that the vertical arrows between $\mathcal{L}$ and $\mathcal{M}$ are identities and the middle vertical arrow a isomorphism of $\mathcal{O}_X$-modules.
QUESTION 1: by $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M}) \cong H^1(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$ the Ext-group is also endowed with structure of a $k$ vector space and I'm asking if there is a nice description how two extension classes in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M}) $ differ from each other /or related to each other if their corresponding elements in $H^1(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$ differ by a multiplication by a scalar $a \in k^*$:
in other words if 
$$0 \to \mathcal{L} \to E_1 \to \mathcal{M} \to 0$$
and 
$$0 \to \mathcal{L} \to E_2 \to \mathcal{M} \to 0$$
are two representers of two extension classes in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ and the vectors $v_{E_1}$ and $v_{E_2} \in H^1(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$ lie on the same line $k \cdot v_{E_1}$:
i.e. there exist a $a \in k^*$ with $v_{E_2}=a \cdot v_{E_1}$, is there a meaningful contruction between $E_1$ and $E_2$ in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ relating them to each other in dependence of $a$?
In other words how the two exact sequences of $E_1$ and $E_2$ are in this case related to each other in sophisticated way reflecting that their corresponding vectors in $v_{E_1}$ and $v_{E_2} \in H^1(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$ are only differ by a scalar.
Or more generally, how the action of $k$ on by scalar multiplication 
$H^1(X, \mathcal{M}\otimes  \mathcal{L}^{\vee})$
can be transfered to an action on 
the exact sequences representing extension classes
from $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$?
QUESTION 2: 
How to see that the $0$  in 
$\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$
(the neutral element of this
Abelian group)
corresponds to the class of splitting extension
$$0 \to \mathcal{L} \to \mathcal{L} \oplus \mathcal{M}
 \to \mathcal{M} \to 0$$
I often saw in comments/ remarks on this issue that
people just say 'that's because the two objects are
canonical' from both viewpoints: in a vector space
as well extension classes.
But I nowhere found a "clean" constructive argument 
why this identification is true diving in explicit
machinery how thegroup elements of the Ext^1
group are identified with extension classes.
 A: Remark. Exact sequence $0 \to L \to E \to M \to 0$ corresponds to $Ext^1(M,L)$, not to $Ext^1(L,M)$.
Q1. $a \in k^\times$ acts on $Ext^1(L,M)$ via pullback along $a:L \to L$ or via pushout along $a: M \to M$.
Q2. There are two options: either one can check that the split sequence is the neutral element for the addition, or that in the long exact sequence
$$
0 \to Hom(L,M) \to Hom(L,L \oplus M) \to Hom(L,L) \to Ext^1(L,M)
$$ 
the element $1_L \in Hom(L,L)$ goes to 0.
A: That is we start with an arbitrary extension $0 \to M \to e_2 \to L \to 0$ represented by the class of the image $\Phi_{e_2}:=\delta(id_L)$ with respect the delta-map in lower row in second diagram below and it's pullback extension $e_2$ in the upper row. Now we want determine that the extension $\overline{e_1}$ is represented by multiplication $a \cdot \Phi_{e_2} =: \Phi_{e_1}$.
We apply $Hom(L,-)$ to diagram
$$
\require{AMScd}
\begin{CD}
0 @>  >> M @>  >> e_1 @>a^{-1} >>  L @> >> 0\\
@VVV  @VVV  @VVV  @VV\cdot{a}V \\
0 @>  >> M @>  >> e_2 @> >>  L @> >> 0 
\end{CD}
$$
and obtain
$$
\require{AMScd}
\begin{CD}
Hom(L, E) @>  >> Hom(L,L) @>\delta  >> Ext(L,M) @> >>  \\
@VVV  @VV\cdot{a}V  @VVV  \\
Hom(L,\overline{E}) @>  >> Hom(L,L) @>\delta  >> Ext(L,M) @> >> 
\end{CD}
$$
That's a diagram of $k$-vector spaces. As you explaned in the answer the extension $e_1$ is forced to be the pullback of $e_2$: i.e. $e_1= a^*e_2$. $k$-linearity and commutativity of the maps imply $a \cdot \Phi_{e_2}=a \cdot \delta(id_L) = \delta(a \cdot id_L) = \Phi_{e_1}$. So $e_1=a e_2$. Is this the correct result of the $k^*$ action by scalar multiplication on $Ext(L,M)$? Or do I have somewhere implemented your hints on my question 1) in wrong way?
