Consider a differentiable map $\Psi: M \rightarrow M$ , where $M$ is a differentiable manifold. Let $x^*$ be a hyperbolic fixed point and suppose that the relative stable and unstable manifold intersect transversally in point $x$. Now in view of the invariance of both stable and unstable manifold, I know the image $\Psi(x)$ of $x$ is an intersection point as well.
My question is: why $\Psi(x)$ is the "third" intersection point and not he "second". In other words, why is true the graph of the left side of the image and the graph at right side is wrong? There are some qualitative and intuitive proof of this fact?
P.S. Is it important stay in a two-dimensional manifold?
Both pictures are correct: There is no second or third.
Actually, there are infinitely many such points (those in the orbit of $x$, that is, $\Psi^n(x)$ with $n\in\mathbb Z$), so you can take any of them as "second". In practical terms it means that you can consider any of these images obtained from your own:
The intersections may come from different orbits, simply because we are considering discrete time and so some complicated behaviors can happen, as in the picture:
The dynamics is still continuous and so close points go to close points, but this does not forbid such behavior "between" $x$ and $\Psi(x)$.
It goes without saying that it would be immensely difficult, if at all possible, to give explicitly a dynamics exhibiting such a behavior, especially the one in the second picture. Of course, if a point is in the intersection, then all its orbit is also in the intersection.
Exactly the same happens in any dimension assumming that you are considering a transverse intersection.