The ultrafilter lemma states that every proper filter on a set X is contained in some ultrafilter on X.
Wikipedia says that in ZFC one can prove the ultrafilter lemma, but in ZF it's not possible to prove it. How is it possible to prove that in ZF it's not possible to prove the ultrafilter lemma?
How can you prove that the theory of fields does not prove the existence of some $x$ such that $x\cdot x=1+1$? You exhibit a field in which this statement is false.
How can you prove that $a\cdot b=b\cdot a$ is not provable from the axioms of a group? You exhibit a group in which this statement is false.
So, how can you prove that the ultrafilter lemma is not provable from the axioms of $\sf ZF$? Well, you exhibit a model of $\sf ZF$ in which the statement is false.
Models of set theory tend to be conceptually "harder" since (1) there are no "easy examples" like in the case of groups, or even $\Bbb N$ or $\Bbb R$ being "easy enough examples" for their respective theories, and (2) since we think of $\sf ZF$ as being kind of the basis of mathematics, at least in some sense, it is hard to think about different models of it, because that means different universes of mathematics, and to some extent different mathematics. And that leap is hard to make.
Setting aside these issues, the easiest way to study models of $\sf ZF$, specifically in the context of relative consistency results, is by Cohen's technique of forcing and von Neumann's and Gödel's approach to inner models. Fortunately, you'll need to combine both of these approaches in order to get consistency results without the axiom of choice.
Putting that technicality aside, the easiest way to make something fail, is to add a counterexample. So, a set without a non-principal ultrafilter, for example. But that can be a bit hard to envision. So instead, we can see what consequences the ultrafilter lemma have, and then see if any of those is easy to negate.
Luckily, the ultrafilter lemma has so many consequences and equivalences that it's a serious contender to the "most useful weak choice principle". The easiest thing we can prove is every set can be linearly ordered. This is best proved using the compactness theorem for first-order logic, which itself is equivalent to the ultrafilter lemma.
Now it's a bit easier to negate that. Namely, adding a set which cannot be linearly ordered. The simplest approach to non-well-orderable sets will generally yield sets which are in fact amorphous, that is, they cannot be split into two disjoint infinite sets. And if $A$ is amorphous, then $A$ cannot be linearly ordered. Let's prove that.
Proposition. Suppose that $A$ is an amorphous set, then $A$ cannot be linearly ordered.
Proof. If $<$ was a linear ordering of $A$, then for each $a\in A$, $A_a\{b\in A\mid b<a\}$ and $A^a=\{b\in A\mid a\leq b\}$ forms a partition of $A$, so exactly one of those is infinite; moreover $\{a\in A\mid A_a\text{ is finite}\}$ and $\{a\in A\mid A^a\text{ is finite}\}$ is itself a partition.
Without loss of generality, we can assume that $A_a$ is finite for all but finitely many $a\in A$, otherwise simply consider the reversed order (or the symmetric argument), and by removing the finite remainder we can assume that $A_a$ is finite for every $a\in A$. Now, note that by linearity and finiteness, each $A_a$ has a unique finite size, and since $A$ is infinite, $\{|A_a|\mid a\in A\}$ is an infinite set of natural numbers in bijection with $A$. So by splitting that set of integers into two infinite subsets, we can get two disjoint infinite subsets of $A$, which is a contradiction to being amorphous. $\square$
Now, you might ask, how do we prove that the existence of an amorphous set is even possible? Well, that requires a lot more technical details. There are numerous technical "blackbox-able" theorems that will tell you that given a structure in some language $\cal L$ with a sufficiently rich automoprhism group, and a sufficiently rich algebra of subsets of our structures, we can create a "copy" of this structure where the algebra of subsets is the power set of our "copy".
Applying this to $\cal L$ being the empty language, our structure an infinite set, the autmorphism group being just all the permutations, and the subsets being the finite and co-finite sets will traditionally satisfy these requirements and result in an amorphous set. If you want to longer version, I recommend starting with Jech's "The Axiom of Choice" book or with Halbeisen's "Combinatorial Set Theory".