Are there counter intuitive interpretations of ZF or ZFC?
This could be along the lines of what you are looking for, although it is only a model for $ZF$ minus the axiom of infinity. Define a binary relation $\in_{\Bbb N}$ on the set of nonnegative integers $\Bbb N$ by $$\forall k,n\in\Bbb N\;\Big(k\in_{\Bbb N}n\Longleftrightarrow\text{the }k\text{-th binary digit of }n\text{ is nonzero}\Big)$$ Then $(\Bbb N,\in_{\Bbb N})$ satisfies all the axioms of $ZF$ except for the axiom of infinity.
This is called the BIT predicate or Ackermann coding.
There is a notion of "standard model" for $\sf ZFC$. But for that we first need to understand two things:
Since set theory includes inherently infinite objects, there is no "natural interpretation" for set theory, like there is for Peano axioms, or like we have developed for $\Bbb R$ over all these years. There is a good reason to call these model "standard", and we'll get to that later.
Since we want to talk about models of set theory, we need to have some notion of set first, otherwise we can't really talk about models. So we will assume to be working within $\sf ZFC$. How can we work inside $\sf ZFC$ without having a model of $\sf ZFC$ first? We can either take it for granted that there is some large universe of sets that happens to obey the laws of $\sf ZFC$; or we can say that we are really writing down definitions and proofs, and we really work in some weak foundational theory, and what we are really proving is that $\sf ZFC$ proves that such and such is true and that we say that such and such definition is called a "standard model".
Yes. This is a big problem to swallow in one bite. If you are uncomfortable with this idea, then you might want to take a year or two to study set theory seriously, and study logic more seriously, and at some point this issue will resolve itself (as do most problems in mathematics: they resolve themselves after sufficient study).
So we are working inside a universe of sets, which happened to satisfy $\sf ZFC$, or some related theory. This means that we already have a notion of $\in$. The one that comes with that universe of sets. Let us call this universe $V$.
So a model of $\sf ZFC$ is a set $M$, in $V$, and some binary relation $E$ on $M$, such that $(M,E)$ satisfy all the axioms of $\sf ZFC$.
Now we can ask, if there is a set $M$ such that $(M,\in)$ (in which case I really mean $\in\restriction(M\times M)$) is a model of $\sf ZFC$. It might be true that such set exists, or it might be false. This will depend on $V$. As it was mentioned by others, $V$ might be such that there aren't any models of $\sf ZFC$. Since if we could prove that there is always such model, then $\sf ZFC$ would prove its own consistency, which we know it can't.
If such $M$ exists, that $(M,\in)$ is a model of $\sf ZFC$ then by Mostowski's collapse lemma, we can "collapse" it, and get a transitive set, namely every element of $M$ will be a subset of $M$. And this means that $M$ really perceives the notion of $\in$ as the universe does. And in that case we say that $M$ is a standard model of $\sf ZFC$.
And since there are standard models, there are non-standard models too. Any model which is not isomorphic to a standard model is called non-standard. The strange thing about non-standard models is that they are never well-founded (otherwise Mostowski's collapse lemma would allow us to make them isomorphic to a standard model). This means that if $(M,E)$ is a non-standard model, then there is a sequence $X=\{x_n\mid n\in\Bbb N\}$ of elements in $M$, such that $x_{n+1}\mathrel E x_n$. How does that sit well with the fact that $M$ satisfies the axiom of foundation? Well, $M$ is aware of the existence of each of the $x_n$'s separately, but it does not know the entire collection of the $x_n$'s. Namely, there is no set in $M$ which represents $X$, or in other words, there is no $Y\in M$ such that $X=\{y\in M\mid y\mathrel E Y\}$.
Non-standard models can be easily obtained from any model by taking ultrapowers or by using compactness. On the other hand, it is possible that there are no standard models, only non-standard models, much like it is possible that there are no models at all.
So what would be a counter-intuitive interpretation of $\sf ZFC$? I'd say that any non-standard model would be. As a result, this means that any interpretation which doesn't know of models of $\sf ZFC$ is also non-standard (since the statement "There is a model of $\sf ZFC$" can be translated to a statement about Gödel numbers, which is absolute between standard models, so if a standard model exists, it must know about models of $\sf ZFC$ itself).
There is a lot more to say about these things. There are delicate points, and common pitfalls that one would implicitly assume are true, but don't have to be. And there are ways to get around those as well. The standard interpretations, if so, would be those which don't have these pitfalls (or at least not the most of them).
One important addition that I'd still like to make, is that given any model of $\sf ZFC$, we can work internally to that model, and now that model is our universe, and we can do anything that $\sf ZFC$ proves possible within that model. So philosophically, you can view mathematics as something in a fixed universe of set theory, or a fixed universe of mathematics, or you can view set theory as something which has many different universes, which are different models of set theory in larger universes (which may, themselves, be models of even larger universes and so on).
This is very late to the party, but I think it's worth noting that there are in fact specific obstacles to finding "natural" nonstandard models of set theories, coming from computability theory. To start with, we have:
There is no computable model of $\mathsf{ZF}$.
(The argument here works for other set theories like $\mathsf{NFU}$ as well, it's quite general.)
That is, there is no computable binary relation $R$ on $\mathbb{N}$ such that $(\mathbb{N},R)\models\mathsf{ZF}$. And things get even worse if we demand "nice" models: there is no hyperarithmetic model of $\mathsf{ZF}$ with no "false natural numbers" (= $\omega$-model), and the situation is even nastier for fully well-founded models (well-founded models are guaranteed to be $\omega$-models, but not conversely).
If we drop the axiom of infinity, things get better per Olivier Begassat's answer above. However, they don't get much better: Tennenbaum's theorem winds up implying that there are no other computable models of $\mathsf{ZFC}$-$\mathsf{Inf}$. (Interestingly, $\mathsf{ZF}$-$\mathsf{Inf}$ is a more complicated beast - see here.)
So while the downward Lowenheim-Skolem theorem tells us that if $\mathsf{ZFC}$ is consistent then it has a very "unintended" model (namely a countable one), we aren't going to find any particularly simple models.