Let $k$ be an algebraically closed field of characteristic $0$, $K$ a function field over $k$, $A=(A,m)$ a discrete valuation ring of rank one with residue field $\kappa_A=A/m$ such that $k \subset A \subset K$ and the fraction field of $A$ is $K$.
Let $\widehat{K}$ be the completion
of $K$ for $A$. More precisely it's the fraction field of $\widehat{A}_m$, the completion of $A$ with respect $m$. Why the completion $\widehat{K}$ is isomorphic to $\kappa_A((T))$?
(Source: Unramified cohomology and rationality problems by Emmanuel Peyre)
Obviously we have a map $k \to \kappa_A$. Let $t$ be the parameter of $A$, ie the generator of $m$. We obtain a map $k[T] \to A, T \to t$, inducing $k[[T]] \to \widehat{A}$ after completion. How to prolonge it to $ \kappa_A[[T]]$?
Take $s\in m$ transcendental over $k$.
$K$ is a finite extension of $k(s)$ so $\widehat{K}$ is a finite extension of $k((s))$.
Let $L/k((s))$ be the maximal unramified subextension of $\widehat{K}/k((s))$. By Hensel lemma we get $L = \kappa_A((s))$.
$\widehat{K}/L$ is a tamely totally ramified finite extension, of degree $n$, so that $\widehat{K}=L((rs)^{1/n})$ for some $r\in \kappa_A^*$.
ie. $$\widehat{K}= \kappa_A(((rs)^{1/n}))\cong \kappa_A((T))$$