My favorite use of ultrafilters is for defining ultralimits. The wikipedia page explains them pretty well, but basically it lets you extend the notion of convergence (of sequences of real numbers, for example) in such a way that every bounded sequence converges, and these limits still respect sums.
You can use ultralimits to make precise statements like "exactly half of the integers are even, and one-third of them are multiples of three". You can do this by defining a "measure" $\mu$ on the integers with the following properties:
$\mu$ is a function from sets of integers to [0,1]
$\mu(\mathbb{Z}) = 1$
$\mu(A \cup B) = \mu(A) + \mu(B)$ for disjoint sets $A,B \subset \mathbb{Z}$
$\mu$ is translation invariant, ie $\mu(A) = \mu(n + A)$ for all $n \in \mathbb{Z}$
This measure measures the "proportion" each subset of $\mathbb{Z}$ takes up. To define it, we'd like to do something like taking the proportion of larger and larger intervals that are part of the set:
$\mu(A) = \lim_{n\rightarrow \infty} \frac{\lvert A \cap [-n,n]\rvert}{2n}$
This works fine for having the set of even numbers be $\frac{1}{2}$, but this limit doesn't always exist (for example, if $A$ is the set of numbers whose first digit is 1).
So the solution: Use an ultrafilter! Pick an ultrafilter, and use the corresponding ultralimit in place of the limit in the above definition of $\mu$. Now it converges for all $A$, since every sequence of numbers in [0,1] converges in an ultralimit! The invariance under translation is pretty easy to see, and the finite additivity follows from additivity of ultralimits. So we've defined something very close to a uniform probability distribution on the integers, and all it took was the axiom of choice.
You can actually do this trick to form such a measure on any group that is amenable, but getting into a discussion of such groups would probably be going a little off-topic here.