From this answer I get that within each cycle you map each element to the one on the right, when taking the product of cycles the one on the right should be performed first, as a typical operator.
Then it would seem to me that
$$(x_1 x_2 x_3)=(x_1 x_3)(x_1 x_2).$$
But according to M. Hall, The Theory of Groups on page 60 of the book, it seems that
$$(x_1 x_2 x_3)=(x_1 x_2)(x_1 x_3).$$
So my question is, are there two different conventions existing for the order of cycle products? Or did someone make a typo? Or are all authors saying the same thing and did I make an interpretation error?
It's no error, it stems from the fact that there are two different ways to compose permutations (and functions, more generally).
Each person/book/article has their own convention about whether functions act on the left or on the right. That is, if you want to apply a function $f$ to a point $x$, do you choose the more universal $f(x)$, or the "backwards" $(x)f$ way that is essentially unique to (some) algebraists?
One benefit of the latter is that $f \circ g$ really means "do $f$ first, then $g$": we would compute $(x)(f \circ g) = ((x)f)g$. It's also a bit nicer for permutations, in some sense. We always work left to right. Anyone choosing this convention is going against the grain, and should be specifying this somewhere in the book.
As a personal anecdote, a few of my algebra professors were students of the group-theorist Martin Isaacs, Isaacs being the biggest proponent of the "functions act on the right" convention (that I'm personally aware of). They preferred his notation, accordingly. In the basic abstract algebra courses, they used the usual "on the left" convention (as do most textbooks, including the one we were using). But then when the courses reached the graduate level, they switched to their more familiar "on the right" style. Because of this slightly-jarring transition, I've been rather aware of the issue. I learned somewhat recently that one of the classic algebra texts also use this "algebraist's" convention; I think it's Hungerford, but I'm not sure (hopefully somebody can remind me).
Now that I think about it, this explains why the following is the second exercise in Isaacs' Algebra:
to show that the two different conventions are equivalent in the sense that they yield isomorphic groups!
See here for the discussion on MathOverflow.