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.



Action on group $\operatorname{Ext}^i(\mathcal{L}, \mathcal{M})$ by scalar multiplication
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.


