[proofplan]
The point is that a short exact sequence ending in a projective module splits. Since $P''$ is projective, the given sequence identifies $P$ with the direct sum $P'\oplus P''$. The equality in $K_0(R)$ then follows from the definition of addition by direct sum, first in the monoid $V(R)$ of finitely generated projective modules and then after passing to its group completion.
[/proofplan]
[step:Split the short exact sequence using projectivity of the quotient]
Since $P''$ is projective and $0 \longrightarrow P' \xrightarrow{i} P \xrightarrow{q} P'' \longrightarrow 0$ is short exact, [citetheorem:8640] applies. Hence the sequence splits. Therefore there exists an $R$-[linear map](/page/Linear%20Map) $s:P''\to P$ such that $q\circ s=\operatorname{id}_{P''}$. From this section we build the explicit direct-sum decomposition. Define the map $\Phi:P'\oplus P''\to P$ by $\Phi(x,y)=i(x)+s(y)$ for every $x\in P'$ and $y\in P''$. We prove that $\Phi$ is an isomorphism of left $R$-modules. To prove injectivity, suppose $\Phi(x,y)=0$ for some $x\in P'$ and $y\in P''$. Applying $q$ gives $y=(q\circ s)(y)=q(i(x)+s(y))=q(0)=0$. Then $i(x)=0$, and injectivity of $i$ gives $x=0$. Thus $\Phi$ is injective. To prove surjectivity, let $p\in P$. Since $q(p-s(q(p)))=q(p)-q(p)=0$, exactness gives $p-s(q(p))\in\ker q=\operatorname{im}i$. Hence there exists $x\in P'$ such that $i(x)=p-s(q(p))$. With $y=q(p)\in P''$, we have $\Phi(x,y)=p$. Therefore $\Phi$ is surjective, and hence $\Phi$ is an isomorphism of left $R$-modules.
[guided]
We use the projectivity of $P''$ exactly to split the surjection $q:P\to P''$. The short exact sequence hypothesis says that $q$ is surjective, that $i$ is injective, and that $\operatorname{im} i=\ker q$. Since $P''$ is projective, [citetheorem:8640] applies to this short exact sequence. It gives an $R$-linear section $s:P''\to P$ of $q$, meaning
\begin{align*}
q\circ s=\operatorname{id}_{P''}.
\end{align*}
From this section we build the explicit direct-sum decomposition. Define the map $\Phi:P'\oplus P''\to P$ by $\Phi(x,y)=i(x)+s(y)$ for every $x\in P'$ and $y\in P''$. This map is $R$-linear because $i$ and $s$ are $R$-linear. We now check directly that it is an isomorphism. Suppose first that $\Phi(x,y)=0$ for some $x\in P'$ and $y\in P''$. Applying $q$ removes the $i(x)$ term because $q\circ i=0$, and it sends $s(y)$ back to $y$ because $q\circ s=\operatorname{id}_{P''}$. Hence $y=(q\circ s)(y)=q(i(x)+s(y))=q(0)=0$. With $y=0$, the equality $\Phi(x,y)=0$ becomes $i(x)=0$. Since exactness at $P'$ says that $i$ is injective, we get $x=0$. Therefore $\Phi$ is injective.
For surjectivity, take an arbitrary element $p\in P$. We want to write $p$ as something from $i(P')$ plus something from the section $s(P'')$. The natural candidate for the $P''$-component is $q(p)\in P''$. Subtracting its lifted copy gives $q(p-s(q(p)))=q(p)-q(s(q(p)))=q(p)-q(p)=0$. Thus $p-s(q(p))\in\ker q$. Exactness at $P$ gives $\ker q=\operatorname{im}i$, so there exists $x\in P'$ with $i(x)=p-s(q(p))$. Setting $y=q(p)$ gives $\Phi(x,y)=i(x)+s(y)=p-s(q(p))+s(q(p))=p$. So $\Phi$ is surjective. Since $\Phi$ is both injective and surjective, it is an isomorphism of left $R$-modules, and therefore $P\cong P'\oplus P''$ as left $R$-modules.
[/guided]
[/step]
[step:Translate the direct sum decomposition into the defining relation in $K_0(R)$]
Let $V(R)$ denote the commutative monoid of isomorphism classes of finitely generated projective left $R$-modules, with addition defined by direct sum. Let $\iota:V(R)\to K_0(R)$ be the canonical monoid homomorphism into its group completion. For a finitely generated projective left $R$-module $M$, write $[M]$ for $\iota([M]_{V(R)})$.
Since $P\cong P'\oplus P''$, the corresponding classes in $V(R)$ satisfy
\begin{align*}
[P]_{V(R)}=[P'\oplus P'']_{V(R)}.
\end{align*}
By the definition of addition in $V(R)$,
\begin{align*}
[P'\oplus P'']_{V(R)}=[P']_{V(R)}+[P'']_{V(R)}.
\end{align*}
Applying the monoid homomorphism $\iota:V(R)\to K_0(R)$ gives
\begin{align*}
[P]=[P']+[P''].
\end{align*}
This is the desired equality in $K_0(R)$.
[/step]
