When I took an introductory model theory course a few years ago, the instructor did not talk about ultrapowers, which seems fair to me, given contemporary model theorists' relative indifference to ultraproducts in general.
On the other hand, ultrapowers ares still sometimes used, e.g., to introduce to students applications of model theory, such as infinitesimals or modal logic (for the latter, see, e.g., Blackburn, de Rijke, Venema's Modal Logic). As far as I know, for these purposes you don't really need ultrapowers; you just need sufficiently saturated elementary extensions, which can easily be obtained by repeated use of compactness.
(A seemingly more recent applications are in operator algebra, but I'm not sure if you really need ultrapowers as opposed to any sufficiently saturated elementary extensions.)
I would imagine that whatever property you want in your elementary extension (e.g., $\kappa$-saturated or omitting a certain type), it is impossible or harder to achieve that with ultrapowers since the latter involves intricate infinite combinatorics.
So my question is: are there situations in which ultarpowers are more desirable than other kind of elementary extensions?
(Note: I understand that there are mainly set-theoretic/combinatorial interests in ultrapowers. I also vaguely understand the uses of ultraproducts; my question is specifically in regard to ultrapowers.)
One pleasant feature of ultrapowers is captured in the slogan "ultraproducts commute with reducts". That is, given two languages $L\subseteq L'$, a collection $(M_i)_{i\in I}$ of $L'$-structures, and an ultrafilter $U$ on $I$, we have $(\prod_I M_i/U)|_L = \prod_I (M_i|_L)/U$ as $L$-structures.
Reformulating this, suppose $M$ is an $L$-structure, and $N = M^I/U$ is an ultrapower of $M$ (in particular, $M\prec_L N$). If we expand $M$ to an $L'$-structure $M'$ by adding new relations or functions, then $N$ has a canonical expansion to an $L'$-structure $N' = (M')^I/U$, and we still have $M'\prec_{L'} N'$.
Moreover, this expansion is totally natural. For example, looking at $\mathbb{R}$ and an ultrapower $\mathbb{R}^*$ (the hyperreals), your favorite function $f\colon \mathbb{R}\to \mathbb{R}$ induces a function $f^*\colon \mathbb{R}^*\to \mathbb{R}^*$, defined by $f^*([a_i]) = [f(a_i)]$, where $[a_i]$ is the equivalence class of the sequence $(a_i)_{i\in I}$ under $U$-equivalence. And we have $(\mathbb{R},f)\prec (\mathbb{R}^*,f^*)$. The same goes for relations: if we add a predicate $Z$ to $\mathbb{R}$ picking out the integers, we have $(\mathbb{R},Z)\prec (\mathbb{R}^*,Z^*)$, where $[a_i]$ satisfies $Z^*$ just in case $U$-almost-all of the $a_i$ are integers.
Yet another way of putting this is that the canonical embedding of $M$ in its ultrapower $N$ doesn't just preserve first-order formulas, but also existential second order formulas: For any tuple $\overline{a}$ from $M$ and new relation symbols $R_1,\dots,R_k$, if $M\models \exists R_1,\dots,R_k\,\varphi(\overline{a})$, where $\varphi$ is a first-order formula in the language $L' = L\cup \{R_1,\dots,R_k\}$, then $N\models \exists R_1,\dots,R_k\,\varphi(\overline{a})$. We get this by expanding $M$ to $L'$ by relations $R_1,\dots,R_k$ witnessing the second-order existential quantifiers, expanding $N$ into an $L'$-structure in the canonical way, and then using $M\prec_{L'} N$. As Noah points out in the comments, the converse does not necessarily hold (the embedding $M\to N$ does not necessarily reflect truth of existential second-order formulas).
I believe that any saturated elementary extension of $M$ (i.e. saturated in its own cardinality) will share this property, and I also believe that it's possible for $M$ to have a $\kappa$-saturated elementary extension of size $\lambda>\kappa$ which fails this property. Unfortunately, I don't have proofs of these facts in mind (can anyone supply proofs/counterexamples?). But in any case, the expansion of an ultrapower is canonical and easy to explain, which is definitely an advantage (especially for people who want to know about nonstandard analysis but don't know much model theory).