Understanding a certain proof about $R$-module homomorphisms and split exact sequences I'm currently reading "Algebra: Chapter 0" by Paolo Aluffi. 
Before I state my problem, I would like to give here the definition of split exact sequences from the book:

A short exact sequence
$0 \to M_1 \to N \to M_2 \to 0$
"splits" if there is a commutative diagram

in which the vertical maps are all isomorphisms.

Now, I'm having trouble with understandg a part of a given proof of the following proposition:

Let $\phi: M \to N$ be an injective $R$-module homomorphism. Then $\phi$ has a left-inverse(as a homomorphism, not a function) if and only if the sequence 
$0 \to M \stackrel{\phi_1}{\rightarrow} N \to coker \ \phi \to 0$
splits.

Here is the proof from the book(I will state only the part I'm having trouble understanding with, that is, the "if" part of the statement):

If the sequence splits, then $\phi$ may be identified with the embedding of $M$ into a direct sum $M \oplus M'$, and the projection $M \oplus M' \to M$ gives a left-inverse of $\phi$.

Well, I'm a bit a lost here:
First of all, why $\phi$ can be identified with the embedding of $M \to M \oplus M'$? According to the definition above, it can be identidied with a surjective function $M'_1 \oplus M'_2$ where $M'_1 \cong M$ and $M'_2 \cong \frac{N}{im \ \phi} \cong coker \ \phi$
 A: Suppose the sequence $0\to M\xrightarrow{\phi}N\to\operatorname{coker}\phi\to0$ splits. Then there exists the diagram
$$\DeclareMathOperator{\coker}{coker}\require{AMScd}
\begin{CD}
0 @>>> M @>\phi>> N @>\pi>> \coker\phi @>>> 0 \\
@. @V{\alpha}V\sim V @V\beta V\sim V @V\gamma V\sim V @. \\
0 @>>> M' @>f>> M'\oplus C' @>g>> C' @>>> 0
\end{CD}
$$
as specified in the definition.
Our task is to find $\psi\colon N\to M$ such that $\psi\phi=1_M$. It is implicit in the definition that the maps in the bottom row are the canonical ones, so there exists $h\colon M'\oplus C'$ such that $hf=1_{M'}$.
Now define $\psi=\alpha^{-1}h\beta$; then
$$
\psi\phi=\alpha^{-1}h\beta\phi=\alpha^{-1}hf\alpha=
\alpha^{-1}\alpha=1_M
$$
The “identification” is exactly considering $f$; it seems that the author is trying to confuse the reader instead of applying the definition.
For proving the converse, you need to build the diagram by exploiting the existence of $\psi\colon N\to M$ with $\psi\phi=1_M$. The diagram is
$$
\begin{CD}
0 @>>> M @>\phi>> N @>\pi>> \coker\phi @>>> 0 \\
@. @V{1_M}V\sim V @V\beta V\sim V @V1_{\coker\phi} V\sim V @. \\
0 @>>> M @>f>> M\oplus\coker\phi @>g>> \coker\phi @>>> 0
\end{CD}
$$
where $\beta(x)=(\psi(x),\pi(x))$. The diagram commutes because $g\beta=\pi$ and, for $x\in M$,
$$
\beta\phi(x)=(\psi\phi(x),\pi\phi(x))=(x,0)=f(x)
$$
A: EDIT: Based on your comment I changed the answer slightly to solve the confusion...
The point is we have an injective map $\phi_1: M\to N$ and we are looking for a map (a retraction) $\rho: N\to M$ such that $\rho\phi_1 = \mathrm{id}_M$. Let $C=\mathrm{Coker}\phi_1$. The exact sequence $0\to M\xrightarrow{\phi_1} N\xrightarrow{\kappa} C\to 0$ is split (where $\kappa$ is the quotient map). So there are $R$-modules $M'_1, M_2'$ such that the diagram you drew is commutative with all the vertical maps isomorphisms. Let us put some names on these isomorphisms $\alpha: M\xrightarrow{\sim} M'_1$, $\beta: C\xrightarrow{\sim} M'_2$ and $\psi: N\xrightarrow{\sim} M'_1\oplus M'_2$.
(This part is new) First of all note that you also have an exact sequence $0\to M\xrightarrow{\imath} M\oplus C\xrightarrow{\pi}C\to 0 $ in addition to $0\to M'_1\xrightarrow{\imath'} M_1'\oplus M_2'\xrightarrow{\pi'}M_2'\to 0 $. Clearly
$$
\imath' \circ \alpha = (\alpha\oplus\beta)\circ \imath, \quad
\beta\circ\pi = \pi'\circ (\alpha\oplus\beta)
$$
This means you can without loss of generality assume that $M'_1=M$, $\alpha=\mathrm{id}_{M}$, $M'_2=C$, $\beta=\mathrm{id}_C$ since 0\to M\xrightarrow{\imath} M\oplus C\xrightarrow{\pi}C\to 0$ also satisfies the conditions you listed in the definition of split exact sequence.
Consider the map $\rho:N\xrightarrow{\psi}M\oplus C\xrightarrow{\pi_M}M$ where the second map is the projection to $M$. The claim is $\rho\phi_1=\mathrm{id}_M$. Note that $\mathrm{Im}(\phi_1)\simeq M$ because $\phi_1$ is injective. Moreover this isomorphism is exactly $\psi$ itself. So take $x\in M$, then $\psi\phi_1(x)=(x,0)$. Hence $\rho\phi_1$ is indeed identity. It is in this sense that $\phi_1$ can be identified with the embedding $M\to M\oplus C$.
