[proofplan]
We use the direct-summand characterization of finitely generated projective modules. Since $P$ is finitely generated projective, it is a direct summand of a finite free $R$-module; localizing the resulting decomposition shows that $S^{-1}P$ is a direct summand of a finite free $S^{-1}R$-module. Finally, localization preserves isomorphisms and finite direct sums, so the assignment on finitely generated projective modules is a monoid homomorphism and hence induces the asserted homomorphism on $K_0$.
[/proofplan]
[step:Regard localization as a functor to $S^{-1}R$-modules]
Set $A := S^{-1}R$. For every $R$-module $M$, let $S^{-1}M$ denote the localization of $M$, regarded as an $A$-module by
\begin{align*}
\frac{r}{s}\cdot \frac{m}{t} := \frac{rm}{st}
\end{align*}
for $r \in R$, $m \in M$, and $s,t \in S$.
If $f:M\to N$ is an $R$-[linear map](/page/Linear%20Map), define
\begin{align*}
S^{-1}f:S^{-1}M&\to S^{-1}N
\end{align*}
by
\begin{align*}
(S^{-1}f)\left(\frac{m}{s}\right)=\frac{f(m)}{s}.
\end{align*}
The defining [equivalence relation](/page/Equivalence%20Relation) for localization shows that $S^{-1}f$ is well-defined and $A$-linear. Thus localization carries isomorphic $R$-modules to isomorphic $A$-modules.
[/step]
[step:Localize finite direct sums explicitly]
Let $M$ and $N$ be $R$-modules. Define an $A$-linear map
\begin{align*}
\Theta_{M,N}:S^{-1}(M\oplus N)&\to S^{-1}M\oplus S^{-1}N
\end{align*}
by
\begin{align*}
\Theta_{M,N}\left(\frac{(m,n)}{s}\right)=\left(\frac{m}{s},\frac{n}{s}\right).
\end{align*}
Define
\begin{align*}
\Psi_{M,N}:S^{-1}M\oplus S^{-1}N&\to S^{-1}(M\oplus N)
\end{align*}
by
\begin{align*}
\Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)=\frac{(tm,sn)}{st}.
\end{align*}
These maps are well-defined because replacing a fraction in a localization only multiplies the numerator by an element of $S$, and the displayed formula uses a common denominator. A direct substitution gives
\begin{align*}
\Theta_{M,N}\circ \Psi_{M,N}=\operatorname{id}_{S^{-1}M\oplus S^{-1}N}
\end{align*}
and
\begin{align*}
\Psi_{M,N}\circ \Theta_{M,N}=\operatorname{id}_{S^{-1}(M\oplus N)}.
\end{align*}
Therefore
\begin{align*}
S^{-1}(M\oplus N)\cong S^{-1}M\oplus S^{-1}N
\end{align*}
as $A$-modules.
[guided]
The point of this step is to make the compatibility with direct sums completely explicit, because this compatibility is what later makes localization descend to $K_0$. Let $M$ and $N$ be $R$-modules. We define
\begin{align*}
\Theta_{M,N}:S^{-1}(M\oplus N)\to S^{-1}M\oplus S^{-1}N
\end{align*}
by
\begin{align*}
\Theta_{M,N}\left(\frac{(m,n)}{s}\right)=\left(\frac{m}{s},\frac{n}{s}\right).
\end{align*}
This map separates the two components after localization.
Conversely, if two localized elements have different denominators, we combine them by passing to a common denominator. Define
\begin{align*}
\Psi_{M,N}:S^{-1}M\oplus S^{-1}N\to S^{-1}(M\oplus N)
\end{align*}
by
\begin{align*}
\Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)=\frac{(tm,sn)}{st}.
\end{align*}
The denominator $st$ lies in $S$ because $S$ is multiplicative. The numerator $(tm,sn)$ lies in $M\oplus N$ because $M$ and $N$ are $R$-modules.
We now verify that these two maps are inverse. Starting from $\left(\frac{m}{s},\frac{n}{t}\right)$, we compute
\begin{align*}
\Theta_{M,N}\left(\Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)\right)=\Theta_{M,N}\left(\frac{(tm,sn)}{st}\right).
\end{align*}
Hence
\begin{align*}
\Theta_{M,N}\left(\Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)\right)=\left(\frac{tm}{st},\frac{sn}{st}\right)=\left(\frac{m}{s},\frac{n}{t}\right).
\end{align*}
Starting from $\frac{(m,n)}{s}$, we compute
\begin{align*}
\Psi_{M,N}\left(\Theta_{M,N}\left(\frac{(m,n)}{s}\right)\right)=\Psi_{M,N}\left(\frac{m}{s},\frac{n}{s}\right).
\end{align*}
Therefore
\begin{align*}
\Psi_{M,N}\left(\Theta_{M,N}\left(\frac{(m,n)}{s}\right)\right)=\frac{(sm,sn)}{s^2}=\frac{(m,n)}{s}.
\end{align*}
Thus $S^{-1}(M\oplus N)\cong S^{-1}M\oplus S^{-1}N$ as $A$-modules.
[/guided]
[/step]
[step:Identify the localization of a finite free module]
For every integer $n\ge 0$, repeated application of the preceding direct-sum isomorphism gives
\begin{align*}
S^{-1}(R^n)\cong (S^{-1}R)^n=A^n
\end{align*}
as $A$-modules. In particular, localization sends finite free $R$-modules to finite free $A$-modules of the same rank.
[/step]
[step:Localize a finite free summand decomposition of $P$]
Since $P$ is finitely generated projective, there exist an $R$-module $Q$, an integer $n\ge 0$, and an $R$-module isomorphism
\begin{align*}
\alpha:P\oplus Q\to R^n.
\end{align*}
Localizing $\alpha$ gives an $A$-module isomorphism
\begin{align*}
S^{-1}\alpha:S^{-1}(P\oplus Q)\to S^{-1}(R^n).
\end{align*}
Using the direct-sum and finite-free identifications from the previous steps, we obtain an $A$-module isomorphism
\begin{align*}
S^{-1}P\oplus S^{-1}Q\cong A^n.
\end{align*}
Thus $S^{-1}P$ is a direct summand of the finite free $A$-module $A^n$. By the direct-summand characterization of finitely generated projective modules, $S^{-1}P$ is finitely generated projective as an $A=S^{-1}R$-module.
[/step]
[step:Show that localization descends to $K_0$]
Let $V(R)$ be the commutative monoid of isomorphism classes of finitely generated projective $R$-modules under direct sum, and let $V(A)$ be the corresponding monoid for $A=S^{-1}R$. Define
\begin{align*}
\ell:V(R)\to V(A)
\end{align*}
by
\begin{align*}
\ell([P])=[S^{-1}P].
\end{align*}
This is well-defined because localization sends isomorphic $R$-modules to isomorphic $A$-modules, and the preceding step shows that $S^{-1}P$ is finitely generated projective whenever $P$ is.
For finitely generated projective $R$-modules $P$ and $P'$, the direct-sum isomorphism gives
\begin{align*}
\ell([P]+[P'])=\ell([P\oplus P'])=[S^{-1}(P\oplus P')].
\end{align*}
Hence
\begin{align*}
\ell([P]+[P'])=[S^{-1}P\oplus S^{-1}P']=\ell([P])+\ell([P']).
\end{align*}
Also $\ell([0])=[0]$. Therefore $\ell$ is a monoid homomorphism. Since $K_0(R)$ and $K_0(A)$ are the group completions of $V(R)$ and $V(A)$, respectively, $\ell$ induces a unique [group homomorphism](/page/Group%20Homomorphism)
\begin{align*}
K_0(R)\to K_0(A)
\end{align*}
sending $[P]$ to $[S^{-1}P]$. Replacing $A$ by $S^{-1}R$ gives the asserted homomorphism
\begin{align*}
K_0(R)\to K_0(S^{-1}R), \qquad [P]\mapsto [S^{-1}P].
\end{align*}
This completes the proof.
[/step]
