You don't say what the context is, but let's suppose at first that the question is asked in the large cardinal context of $\kappa$-complete ultrafilters on a measurable cardinal $\kappa$, which is a context where one often considers such questions.

The first thing to say is that there may be no such example, even when there is a measurable cardinal. For example, in the canonical inner model $L[\mu]$ with one measurable cardinal $\kappa$, there is exactly one normal measure $\mu$ on $\kappa$, and every $\kappa$-complete ultrafilter on $\kappa$ is Rudin-Kiesler equivalent to a finite power $\mu^n$ of $\mu$. In particular, although these powers do indeed form an increasing chain $$\mu^1\lt_{RK}\ \mu^2\lt_{RK}\ \mu^3\lt_{RK}\ \cdots$$ in the Rudin-Kiesler order, there can be no measure on top of this chain as you request. In short, it is consistent with a measurable cardinal that the height of the Rudin-Kiesler order is $\omega$.

From a larger large cardinal assumption, however, one can achieve higher Rudin-Kiseler ranks. For example, if $\kappa$ is $\kappa+2$-strong, then there must be ultrafilters $U$ on $\kappa$ with Rudin-Kiesler rank $\omega$, giving rise to the situation of your question. To see this, let $j:V\to M$ with $V_{\kappa+2}\subset M$ and we may assume $M^\kappa\subset M$. Let $X_1=\{j(f)(\kappa)\mid f:\kappa\to V\}$ be the seed hull of $\kappa$, which is elementary in $M$ and collapses to the ultrapower $j_1:V\to M_1$ of the induced normal measure $U_1$ generated by $j$. Consider the first ordinal missing from $X_1$, call it $\delta_1$, and let $X_2$ be the see hull $\{j(f)(\delta_0,\delta_1)\mid f:\kappa^2\to V\}$, using $\delta_0=\kappa$, and similarly define $\delta_n$ for each $n$. The ultrafilter $U_n$ induced by $A\in U_n\iff (\delta_0,\ldots,\delta_{n-1})\in j(A)$ is Rudin-Kiesler below $U$, and they form an increasing chain as desired. The chain is strictly increaing precisely because no $X_n$ can exhaust all of $M$ below $j(\kappa)$.

In the general case, or even in the case of ultrafilters on $\omega$, it remains easy to build increasing chains in the Rudin-Kiesler order. For example, the successive products of a fixed ultrafilter form a strictly increasing chain $$U\lt_{RK} U^2\lt_{RK} U^3\lt_{Rk}\cdots$$
To place an ultrafilter $U$ on top, however, takes some more delicate work (and I see now that Andreas Blass has posted an answer indicated how to undertake that delicate work).