If the exact sequence of left $A$-modules $$0\xrightarrow{}F'\xrightarrow{u}F\xrightarrow{v}F''\xrightarrow{}0$$ splits, the sequence $$0\xrightarrow{}\text{Hom}(E,F')\xrightarrow{\bar{u}}\text{Hom}(E,F)\xrightarrow{\bar{v}}\text{Hom}(E,F'')\xrightarrow{}0$$ is exact and splits, where $\bar{u}=\text{Hom}(1_E,u)$ and $\bar{v}=\text{Hom}(1_E,v)$.
The assertion follows from the fact that $\text{Hom}(E,F')\oplus\text{Hom}(E,F'')$ is canonically identified with $\text{Hom}(E,F'\oplus F'')$ by means of the mapping $\text{Hom}(1_E,j')+\text{Hom}(1_E,j'')$, where $j':F'\rightarrow F'\oplus F''$ and $j'':F''\rightarrow F'\oplus F''$ are the canonical injection.
To show that the sequence is exact and splits, I need to show the following: (1) injectivity of $\bar{u}$, (2) $\text{Im}(\bar{u})=\text{Ker}(\bar{v})$, (3) surjectivity of $\bar{v}$ and (4) the existence of a linear section of $\bar{v}$ (or a linear retraction of $\bar{u})$. (1) and (2) follow from the fact that $$0\xrightarrow{}F'\xrightarrow{u}F\xrightarrow{v}F''$$ is an exact sequence. Since the first sequence splits, there exists an $A$-linear mapping $s:F''\rightarrow F$ such that $v\circ s=1_{F''}$. Thus $$\text{Hom}(1_E,v)\circ\text{Hom}(1_E,s)=\text{Hom}(1_E,v\circ s)=\text{Hom}(1_E,1_{F''}).$$ Hence, $\text{Hom}(1_E,v)$ has a section: i.e. it's surjective. Therefore the second sequence splits and is exact.
Now, I don't understand how my proof relates to the author's suggested method. Can someone please explain to me how the canonical identification proposed also proves the claim?