Let $E'$ and $E''$ be a vector bundles over a complex manifold $X$. An extension of $E''$ by $E'$ is a an exact sequence of vector bundles
$$ 0 \longrightarrow E'\longrightarrow E \longrightarrow E'' \longrightarrow 0. $$
Two extensions of $E''$ by $E'$ are equivalent if we have an isomorphism of exact sequences
$$ 0 \longrightarrow E'\longrightarrow E_1 \longrightarrow E'' \longrightarrow 0, $$
and
$$ 0 \longrightarrow E'\longrightarrow E_2 \longrightarrow E'' \longrightarrow 0. $$
I'm looking for a proof for the following proposition
The equivalence classes of extensions of $E''$ by $E'$ are in one-to-one correspondence with the elements of $H^1(X,\operatorname{Hom}(E'',E'))$, the trivial extension corresponding to the zero element.
I have not seen these extensions defined in pretty much any book on complex/differential geometry and cannot find a proof for the above result. If someone here has knowledge on where these things are discussed I would appreciate it.
In terms of definitions, you can find definitions in any book on homological algebra, (e.g. Weibel's book) though you probably need to be comfortable with what a sheaf is and how to associate a locally free sheaf of $\mathcal O_X$-modules to a vector bundle. In terms of computation, Cech cocycles are probably the most concrete way to understand what an element of either side "looks like".
Note that if $0 \to E' \to E \to E''\to 0$ is a short exact sequence of vector bundles, then if $\mathcal E,\mathcal E',\mathcal E''$ are the corresponding locally free sheaves of sections of $E$,$E'$ and $E''$ respectively, then $0 \to \mathcal E' \to \mathcal E \to \mathcal E'' \to 0$ is a short exact sequence of sheaves. In this way, short exact sequences of vector bundles exactly correspond to short exact sequences of the associated locally-free sheaves.
For any two vector bundles $E'$ and $E''$ you can use the (multi)-linear algebra constructions on each fibre to form vector bundles $E'\otimes E''$ and $\mathcal Hom(E',E'')$, where I use the calligraphic $\mathcal H$ to distinguish the vector bundle from the vector space of maps from the bundle $E'$ to $E''$. These operations are called the "internal" Hom and tensor product for (the category of) vector bundles on $X$.
In contrast, $\text{Hom}(E',E'')$ is the vector space of maps of bundles. In the complex analytic setting, for example, these are holomorphic maps from the total space of $E'$ to the total space of $E''$ which are the identity when restricted to $X$ (viewed as the zero section) and which are linear on the fibres of the bundles. The relationship between the two is straight-forward though -- $\mathrm{Hom}(E',E'')$ is just the space of global sections $\Gamma(X,\mathcal Hom(E',E''))$ of the $\mathcal Hom$-vector bundle (or $\mathcal Hom$-sheaf, since this is defined by requiring its sections over an open set to be space of sections to the bundle map). Similar considerations show that there is a natural identification $\mathbb C_X\otimes E' = E'$.
Now if $\mathbb C_X$ denotes the trivial rank 1 vector bundle on $X$ (that is, $\mathbb C_X = X\times \mathbb C$, then $\phi\in \text{Hom}(\mathbb C_X, E')$ is completely determined by its value on $1 \in \mathbb C$, and $\phi(1)$ is a global section of $E'$, so there is a natural identification $\text{Hom}(\mathbb C_X,E') = \Gamma(X,E')$.
Using this and an "internal" version of tensor-Hom adjunction, which is standard (but perhaps not quite obvious) you can establish the identity relating $\text{Ext}^1$ to $H^1$ of the internal Hom: The identification you ask about can be established as follows: $$ \begin{split} \text{Ext}^1(E'',E')& =\text{Ext}^1(\mathbb C_X \otimes E'',E') = \text{Ext}^1(\mathbb C_X, \mathcal Hom(E'',E'))\\ & \text{R}\mathcal Hom^1(\mathbb C_X,\mathcal Hom(E'',E')) = \text{R}\Gamma^1(X,\mathcal Hom(E'',E')) \\ &=H^1(X,\mathcal Hom(E'',E')). \end{split} $$ as required. Here you can do the calculation using the vector bundle or their associated locally-free sheaves, because it can be checked that the vector bundle $\mathcal Hom(E'',E')$ is that associated to the locally-free sheaf $\mathcal Hom(\mathcal E'',\mathcal E')$ (this assertion implicitly includes the assertion that the $\mathcal Hom$-sheaf is locally free!) and similarly for $E''\otimes E'$ and $\mathcal E''\otimes \mathcal E'$.