We start by a simple observation which is true for any totally ordered set, not only for well-ordered sets.
Let $(X,\le)$ be a linearly ordered set. We will use the notation $X_a=\{x\in X; x<a\}$ pre $a\in X$. Notice that these sets are initial segments of $X$.
Observation. Let $(X,\le)$ be a linearly ordered set and let $X'=\{X_a; a\in X\}$. Then the function $f\colon X\to{X'}$ defined as
$$f(a)=X_a$$
is an isomorphism between the ordered sets $(X,\le)$ and $(X',\subseteq)$.
Proof. We can see immediately that $f$ is surjective. If $a<b$ then $a\in X_b$ and $a\notin X_a$, so $f(a)=X_a\ne X_b=f(b)$. The case $b<a$ is symmetric. So it is also injective.
The map $f$ is monotone: If $a \le b$ then $f(a)=\{x\in X; x<a\} \subseteq \{x\in X; x<b\}=f(b)$. (It suffices to notice that
$x<a$ and $a\le b$ implies $x<b$ by transitivity.)
To see that also $f^{-1}$ is monotone, we notice that $X_a\subseteq X_b$ implies $a\le b$. (If we had $a>b$, then $b\in X_a\setminus X_b$, which contradicts $X_a\subseteq X_b$.) $\hspace{1cm}\square$
Side remark. Notice that we have used linearity of $\le$ in the above proof. The version of the above observation with $\{x\in X; x\le a\}$ instead of $X_a$ is true also for partially ordered sets and it implies that every partially ordered set is isomorphic to some system of sets ordered by inclusion (in fact, to a subset of $(\mathcal P(X),\subseteq)$.) See, for example, Theorem 1.11 in S. Roman: Lattices and Ordered Sets or this question.
How does this relate to ordinals? We can view ordinals as representatives or order types of well-ordered sets. (At least if we approach them in the naive rather than axiomatic way.) We can define inequality between ordinals via initial embeddings of the corresponding well-ordered sets. (I.e., ordinal type of $(A,\le)$ is less or equal to ordinal type of $(B,\preceq)$ iff $A$ is isomorphic to an initial segment of $B$. This seems to be the natural way how to compare ordinals/well-ordered sets.)
So the above lemma shows that a well-ordered set $X$ is isomorphic to the set of proper initial segments of $X$. Ordinal types of these initial segments are precisely the ordinals lesser than the ordinal type of $X$.
There are two ways in which well-foundedness can fail.
Internally, where the model knows that it fails, i.e. the axiom of foundation is false in the model.
Externally, where the model thinks that the axiom of foundation is true, but we know from outside the model that the relation is not really well-founded.
In the latter case you do have a rank function, since the rank function is internal, and there are no decreasing sequences of ordinals in that model. But externally there is a decreasing sequence nonetheless.
This is akin to non-standard models of $\sf PA$. The model is not well-ordered, and we know that, but the induction axioms are equivalent to stating that any definable set which is not empty has a least element, so this is not something the model knows about.
Note that in the context of ultrapowers, the models are all models of $\sf ZFC$, or a significant fragment thereof which contains the axiom of foundation, and therefore they are all going to fall into the second category.
Consider, if you will, a free ultrafilter $U$ on $\omega$, and consider $V^\omega/U$. This is a definable class, with $E$ denoting its membership relation also being definable. This is an ultrapower of the universe. But it is not well-founded, externally, since $\omega^\omega/U$ is already not well-ordered.
(In the former case you will have a witness that is not well-founded, yes. And the ordinals may or may not be well-founded.)
Best Answer
If the sequence is decreasing, it has to be finite, since a well-ordering does not have any infinitely decreasing sequences. But that means that we have to stop somewhere, and if you take "decreasing" to mean "continue to decrease as long as you can" that means that the last element of the sequence must be the minimum, i.e. $0$.