[proofplan]
We use the lifting property in the definition of a projective module to lift the identity map on $P''$ through the surjection $p:P\to P''$. This produces a section $s:P''\to P$ of $p$. The existence of such a section lets us decompose every element of $P$ uniquely as a sum of an element of $\ker p$ and an element of $s(P'')$. Finally, exactness identifies $\ker p$ with $P'$ via the injective map $i$, giving the desired isomorphism $P\cong P'\oplus P''$.
[/proofplan]
[step:Lift the identity map through the surjection $p$]
Since the sequence is exact, the homomorphism
\begin{align*}
p:P\to P''
\end{align*}
is surjective. Since $P''$ is projective, the defining lifting property of projective modules applied to the surjection $p:P\to P''$ and to the identity homomorphism
\begin{align*}
\operatorname{id}_{P''}:P''\to P''
\end{align*}
gives an $R$-linear homomorphism
\begin{align*}
s:P''\to P
\end{align*}
such that
\begin{align*}
p\circ s=\operatorname{id}_{P''}.
\end{align*}
Thus $s$ is a section of $p$, and the short exact sequence splits at the level of the quotient map.
[guided]
The only hypothesis not coming from exactness is the projectivity of $P''$, so we use it immediately. The definition of a projective module says that whenever an $R$-[linear map](/page/Linear%20Map) $q:M\to N$ is surjective and an $R$-linear map $f:P''\to N$ is given, there is an $R$-linear map $\widetilde f:P''\to M$ with
\begin{align*}
q\circ \widetilde f=f.
\end{align*}
In the present situation, exactness of
\begin{align*}
0 \longrightarrow P' \xrightarrow{i} P \xrightarrow{p} P'' \longrightarrow 0
\end{align*}
implies that $p:P\to P''$ is surjective. We take $q=p$, $M=P$, $N=P''$, and
\begin{align*}
f=\operatorname{id}_{P''}:P''\to P''.
\end{align*}
Projectivity of $P''$ therefore gives an $R$-linear homomorphism
\begin{align*}
s:P''\to P
\end{align*}
such that
\begin{align*}
p\circ s=\operatorname{id}_{P''}.
\end{align*}
This equation is exactly the statement that $s$ is a right inverse, or section, of $p$. It means that every element $y\in P''$ can be lifted to the specific element $s(y)\in P$, and applying $p$ to that lift recovers $y$.
[/guided]
[/step]
[step:Decompose $P$ as $\ker p$ plus the image of the section]
Let
\begin{align*}
s(P'')=\{s(y):y\in P''\}\subset P
\end{align*}
denote the image submodule of $s$. We first show that
\begin{align*}
P=\ker p+s(P'').
\end{align*}
For any $x\in P$, define
\begin{align*}
y=p(x)\in P''.
\end{align*}
Then
\begin{align*}
x=\bigl(x-s(p(x))\bigr)+s(p(x)).
\end{align*}
The second summand belongs to $s(P'')$ by definition. For the first summand, $R$-linearity of $p$ and the identity $p\circ s=\operatorname{id}_{P''}$ give
\begin{align*}
p\bigl(x-s(p(x))\bigr)=p(x)-p(s(p(x)))=p(x)-p(x)=0.
\end{align*}
Hence $x-s(p(x))\in\ker p$, proving $P=\ker p+s(P'')$.
Next we show that this sum is direct. If $z\in \ker p\cap s(P'')$, then there exists $y\in P''$ such that $z=s(y)$, and also $p(z)=0$. Therefore
\begin{align*}
0=p(z)=p(s(y))=y.
\end{align*}
Thus $y=0$, so $z=s(0)=0$. Hence
\begin{align*}
\ker p\cap s(P'')=\{0\}.
\end{align*}
Therefore
\begin{align*}
P=\ker p\oplus s(P'').
\end{align*}
[/step]
[step:Identify the two direct summands with $P'$ and $P''$]
Let
\begin{align*}
\operatorname{im} i=\{i(x'):x'\in P'\}\subset P
\end{align*}
denote the image submodule of $i$. Exactness at $P$ gives
\begin{align*}
\operatorname{im} i=\ker p.
\end{align*}
Exactness at $P'$ gives that
\begin{align*}
i:P'\to P
\end{align*}
is injective. Therefore the corestriction
\begin{align*}
i_{\ker}:P'\to \ker p
\end{align*}
defined by
\begin{align*}
i_{\ker}(x')=i(x')
\end{align*}
for $x'\in P'$ is an $R$-linear isomorphism.
Also, the section
\begin{align*}
s:P''\to s(P'')
\end{align*}
is an $R$-linear isomorphism onto its image. Indeed, it is surjective onto $s(P'')$ by definition, and if $s(y)=0$, then applying $p$ gives
\begin{align*}
y=p(s(y))=p(0)=0.
\end{align*}
Thus $s$ is injective.
Combining these identifications with
\begin{align*}
P=\ker p\oplus s(P'')
\end{align*}
gives an $R$-linear isomorphism
\begin{align*}
P'\oplus P''\to P
\end{align*}
defined by
\begin{align*}
(x',y)\mapsto i(x')+s(y).
\end{align*}
Hence
\begin{align*}
P\cong P'\oplus P''
\end{align*}
as left $R$-modules.
[/step]
[step:Apply the splitting to the finitely generated projective case]
If $P'$, $P$, and $P''$ are finitely generated projective left $R$-modules, then in particular $P''$ is projective. The preceding steps apply without any additional hypothesis on $P'$ or $P$, and therefore give
\begin{align*}
P\cong P'\oplus P''.
\end{align*}
This proves the stated special case and completes the proof.
[/step]
