Let me give a low-brow answer to the question, and begin with my earlier answer (which got a couple of downvotes, so you have to take it with a pinch of salt). There I discussed quadratic extensions.

I'm assuming that your base field $K$ is a finite extension of $\mathbb{Q}_p$ or of $\mathbb{F}_p((\pi))$, where $p$ is a prime number and $\pi$ is transcendental. (Very little will change if you allow $K$ to be a field complete for a discrete valuation with perfect residue field.) Let $k$ be the residue field of $K$.

Finite extensions $L|K$ can be *unramified*, *(at worst) tamely ramified* or *wildly ramified.*. The three cases correspond to $e=1$, $\operatorname{gcd}(e,p)=1$, $p|e$, where $e$ is the ramification index of $L|K$. There are uniquely determined subfields $K\subset L_0\subset L'\subset L$ such that $L_0|K$ is unramified, $L'|L_0$ is totally but tamely ramified, and $L|L'$ is totally ramified of degree $p^s$ for some $s\in\mathbb{N}$, so it is wildly ramified if $s>0$. (For me $0\in\mathbb{N}$; I want it to be an additive monoid.)

Unramified extensions can be completely understood in terms of extensions of the residue field $k$. When $k$ is finite as here, there is only one extension in each degree $n$, and it is obtained by adjoining a primitive $(q^n-1)$-th root of $1$, where $q=\operatorname{Card}(k)$. It follows that the maximal unramified extension of $K$ is obtained by adjoining primitive roots of $1$ of order prime to $p$.

Tamely ramified extensions are only slightly more complicated. It is not hard to show that if $L|K$ is totally but tamely ramified of degree $n$, then $L=K(\root n\of\varpi)$ for some uniformiser $\varpi$ of $K$, and not hard to determine when two uniformisers $\varpi$ and $\varpi'$ give the same extension. See for example Lecture 18 in my online notes arXiv:0903.2615. As shown there, the maximal tamely ramified extension $T|K$ is obtained by adjoining $\root n\of1$ and $\root n\of\varpi$ for all $n>0$ prime to $p$, where $\varpi$ is a fixed uniformiser of $K$. This allows you to write a simple presentation for the profinite group $\operatorname{Gal}(T|K)$ in which the generators have some *arithmetic significance.*

(If your base field had been $\mathbb{C}((t))$, all whose finite extensions are totally but tamely ramified, you would have been able to conclude at this point that an algebraic closure is obtained by adjoining an $n$-th root of $t$ for every $n>0$.)

What are all totally ramified extensions $L|K$ of degree $p^s$ ? Very little is known about the question. It is easy to see that there are only finitely many $L$ in the mixed characteristic case, and infinitely many $L$ in the equicharacteristic case, but there are exactly $p^s$ extensions when counted properly.

As Pete says, the *abelian* ones are given by local class field theory (in terms of index-$p^s$ subgroups of $K^\times$). (But even for $s=1$ there are extensions which are not galoisian, and I don't know them all.) The exponent-$p$ ones can be understood not only in terms of Class Field Theory, but also Kummer Theory or Artin-Schreier Theory.

*The question is, loosely put, what is known about wild ramification?*

The answer is, loosely put, not much. (Joking, eh ?)

**Addendum** (26/2/2010). Yesterday I came across a recent theorem of Abrashkin which says, roughly speaking, that if you know all the wildly ramified extensions (along with their filtration), then you know the local field.

More precisely, let $K$ be a local field of residual characteristic $p$, and $P|K$ the maximal pro-$p$-extension of $K$ --- the compositum of all $p$-extensions of $K$. The profinite group $G=\operatorname{Gal}(P|K)$ comes with the ramification filtration (in the upper numbering).

If $K',P',G'$ is another such triple, and if $\varphi:K\to K'$ is an isomorphism of local fields, then it induces an isomorphism of filtered pro-$p$-groups $G\to G'$.

Abrashkin's theorem says that, conversely, every isomorphism of filtered pro-$p$-groups $G\to G'$ comes from an isomorphism $K\to K'$ of local fields. In other words, the local field $K$ is completely determined by the filtered pro-$p$-group $G$. See Theorem A in his recent paper

This is a refinement of an earlier theorem of Mochizuki, who worked with $\operatorname{Gal}(\tilde K|K)$, where $\tilde K$ is a separable closure of $K$.