Intuition of non-concretizable categories?

I don't think that any such example exists, and for good reason. The failure of concretizability is a subtle phenomenon that has nothing to do with abstract algebraic ideas and is instead entirely about size issues (that is, the distinction between sets and proper classes).

To illustrate this last point, if you ignore size issues, then every category is concretizable. Given a category $C$, you can get a faithful functor to sets by sending each object $a$ to the set $F(a)$ of all morphisms to $a$. Given a morphism $a\to b$, you get a map $F(a)\to F(b)$ by taking any morphism to $a$ and just composing it with the morphism $a\to b$. (Note that this construction is exactly the generalization of the usual proof of Cayley's theorem on groups to general categories.)

Now, this doesn't actually work in general if $C$ is a large category, since the "set" $F(a)$ of all morphisms to $a$ may actually be a proper class. But this does show that any small category is concretizable, and any failure of concretizability has to be about size issues. In other words, it has nothing do with sets and their structure per se (any category would be concretizable if you were allowed to use sufficiently large "sets"). Rather, it must have to do with the structure of the category being in some sense "too large" to embed in the category of small sets.

In a category with finite limits being concretizable is equivalent to the fact that RegSub(A) (regular subobjects) is a set for any object A.

This appears in the paper “Concreteness” by Freyd.

Of course this occours just very often, thus a non concretizable category must have “very huge” objects.

For example, a category with a generator is always concrete, this is the case of all the categories you have in mind.

To get natural counterexample to concreteness you can consider the category of presheaves over a big category. The fact that it is not locally small implies that it is not concretizable (in this case).

Isbell's example of a (locally small) non-concretizable category, although arguable artificial as Fosco says in his answer, is fairly simple and it is quite easy to see concretely (sorry!) what makes it work.

The category has two classes $\left\{A_\alpha\right\}$ and $\left\{B_\alpha\right\}$ of objects indexed by the same proper class, and two more objects $X$ and $Y$.

There are morphisms $A_\alpha\to X$, $A_\alpha\to Y$, $X\to B_\alpha$ and $Y\to B_\alpha$ for each $\alpha$, and the only morphisms are the identity morphisms and compositions of these, with the compositions $A_\alpha\to X\to B_\alpha$ and $A_\alpha\to Y\to B_\alpha$ equal (but not the compositions $A_\alpha\to X\to B_\beta$ and $A_\alpha\to Y\to B_\beta$ for $\alpha\neq\beta$).

This is a locally small category, with at most two morphisms between any pair of objects.

Suppose $F$ were a faithful functor from this category to sets, and for a given $\alpha$ let $f_\alpha:F(X)\amalg F(Y)\to F(B_\alpha)$ be the function induced by $X\to B_\alpha$ and $Y\to B_\alpha$.

There is only a set of equivalence relations on $F(X)\amalg F(Y)$, so for some $\alpha\neq\beta$ the equivalence relation induced by $f_\alpha$ (i.e., $\left\{(s,t)\in F(X)\amalg F(Y)\middle\vert f_\alpha(s)=f_\alpha(t)\right\}$) must be the same as that induced by $f_\beta$.

But then, since $F(A_\alpha)\to F(X)\to F(B_\alpha)$ is equal to $F(A_\alpha)\to F(Y)\to F(B_\alpha)$, we must also have that $F(A_\alpha)\to F(X)\to F(B_\beta)$ is equal to $F(A_\alpha)\to F(Y)\to F(B_\beta)$, contradicting faithfulness.