 Solution

For a nonzero Laurent series let ${\mathrm{ord}}_t$ be its least exponent. Choose $|x|=p^{-{\mathrm{ord}}_t(x)}$ and $|0|=0$. This is a multiplicative Non-Archimedean absolute value, and its valuation ring is ${\mathbb{F}}_p[[t]]$.

TakeIf ${\gamma }_1,\dots ,{\gamma }_f$ is an ${\mathbb{F}}_p$-basis of ${\mathbb{F}}_{p^f}$, the elements ${\gamma }_i{u}^j$, $1\le i\le f$, $0\le j<e$, form a $K$-basis of $L$. To see this, group the exponents of each Laurent series modulo $e$ and then expand its coefficients in the constant-field basis. Uniqueness of those two expansions proves linear independence. Thus $[L:K]=ef$.

Its valuation ring is ${\mathbb{F}}_{p^f}[[u]]$, with maximal ideal $(u)$ and residue field ${\mathbb{F}}_{p^f}$. Moreover ${\mathrm{ord}}_u(t)=e$, so the ramification index is $e$ and the residue degree is $f$. The unique extending absolute value is explicitlyThis restricts to the chosen absolute value because ${\mathrm{ord}}_u(x)=e{\mathrm{ord}}_t(x)$ on $K$. The prescribed ramification and residue degree for a Laurent series field construction works even when $p\mid e$; separability is not required in the question.