Why the associative axiom of groups not stated as f * ( g * h ) * = ( f * g ) * h = ( f * h) * g? I am reading Alan Beardon's "Algebra and Geometry" and it says:

(2) for all f, g and h in G,  f * (g * h) = (f * g) * h.
(2) is called the associative law, and it says that f * g * h is uniquely defined regardless of which two operations * we choose to do first. The point here is that * only combines two objects at a time, and we have to apply it twice (in some order) to obtain  f * g * h. There are exactly two ways to do so and the two must always yield the same results.

The issue I have is that I do not see how is it that there are only two ways to combine two elements out of three. For as I understand I could make six different combinations (if I do not assume commutative property):
(1) ( f * g ) * h
(2) ( f * h ) * g
(3) ( g * f ) * h
(4) ( g * h ) * f
(5) ( h * g ) * f
(6) ( h * f ) * g
And even if I assume that there is a commutative property, that I have not seen stablished in the definition so far, there would still be three different combinations:
(1) ( f * g ) * h
(2) ( f * h ) * g
(3) ( g * h ) * f
Can you please explain to me how is it that there are only two ways to apply a binary operation over three elements and thus defining the associative axiom shall be expressed as  f * (g * h) = (f * g) * h and not as ( f * g ) * h = ( f * h ) * g = ( g * h ) * f ?
 A: First, I am a little confused regarding your question: you first point out $6$ different ways to combine $f$, $g$, and $h$. You then say that even if you assume the commutative property there would be three left ... but then when you propose to define the associative property differently, why do you pick only two of those three, and add a different one? Indeed, do you see how weird your proposed definition is, with one of the terms having the parentheses in a different place than the other two?
Anyway, just focusing on two of those: no, we do not have $(f*g)*h=(f*h)*g$ as part of the associative property, because that would require a commutative property as well, and that is something different from the associative property.
I think you are confusing the "ways in which we can perform the order of the operations" and the "ways in which we can order the operands".
Associativity is about the order of the operations. That is, if you have $f * g * h$, then we have two choices for the order of the operations: do we first apply the left *, and then the right - this would be $(f * g) * h$ - or do we first apply the right *, and then the left - which would be $f * (g * h)$? Well, if the * is associative, then it turns out that the order of the operations doesn't matter, i.e. that $f *(g*h) = (f *g)*h$.
You, however, seem to be worrying about the number of ways in which you can order the operands ... and yes, given three operands $f$, $g$, and $h$, you can have many ways to order them, as you point out. However, whether that always gives the same result is not what the associative property is about ... the order of the operands is what the commutative property is about.
A: The thing is, when we write $a\cdot b\cdot c$, you can't first evaluate $a\cdot c$ and then compute $ac \cdot b$, because we don't know if the binary operation is commutative or not. So you don't have six possible way to compute $abc$, you 
only have two, which are $(ab)c$ and $a(bc)$.
A: The definition of associative requires the order to stay the same, without commutative law.  In your first set $(f*g)*h\ne (g*f)*h$, unless $f$ and $g$ are commutative.  Also $(g*h)*f\ne f*(g*h)$ without commutative.  For your second question, those combination are derivable from using both associative and commutative together, so they are not needed for the definition of associative.
A: Groups abstract the idea of transformations that can be applied to some object.  Group elements are individual transformations, and the group operation represents performing the transformations in sequence.
Importantly, group operations are typically not commutative.  If $L$ means "turn 90 degrees left" and $F$ means "walk forward one foot," then $(LF)$ and $(FL)$ are different operations.
To take another example, we have e.g.
$\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix}2 & 3 \\ 0 & 3 \end{pmatrix},$
but
$\begin{pmatrix}2 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix}2 & 2 \\ 0 & 3 \end{pmatrix}.$
On the other hand, obviously $(fg)h = f(gh)$ if $f$, $g$, and $h$ are transformations and the operation is applying them in sequence, because $f(gh)$ means "first do $f$, then (do $g$, then do $h$)," whereas $(fg)h$ means "(first do $f$, then do $g$), then do $h$," and these are of course the same thing.  In other words, a nonassociative operation cannot correspond to combining transformations by applying them in sequence.
A: The author says that there are two ways to understand $f * g * h$ specifically. He never states that there are "two ways to combine two elements out of three" in general. 
((By the way, there are in fact 12 possibilities for 3 elements and 2 identical operators.))
