Localization of the polynomial ring at a prime ideal modulo maximal ideal is isomorphic to polynomial ring modulo prime ideal. Let $p \in K[T]$ irreducible, s.t. $\text{LC}(p) = 1$. Then
$$ K[T]/(p) \cong K[T]_{(p)}/pK[T]_{(p)}.$$
What I have is:
\begin{align*}
&K[T] \hookrightarrow K[T]_{(p)} \text{ and } K[T]_{(p)} \twoheadrightarrow K[T]_{(p)}/pK[T]_{(p)}\\
& \implies K[T] \overset{f}\longrightarrow K[T]_{(p)}/pK[T]_{(p)}
\end{align*}
The kernel of $f$ is $(p)$ so
$$ K[T]/(p) \cong \text{im} f = \left\{g + pK[T]_{(p)} \mid g \in K[T] \right\} \overset{?} = K[T]_{(p)}/pK[T]_{(p)}.$$

Let $\overline{\frac{r}{s}} \in K[T]_{(p)}/(p)K[T]_{(p)}$. Then
\begin{align*}
   p \not | s &\implies \exists a,b \in K[T]: ap + bs = 1\\
& \implies bs = 1 - ap\\
& \implies f(bs) = \overline{bs} = 1 + (p)K[T]_{(p)}\\
& \implies f(rb) =f(r) f(b) \equiv \overline{\frac{r}{s}} 
\end{align*}
Hence
$$ \text{im } f = K[T]_{(p)}/(p)K[T]_{(p)}.$$
