I am trying to get an intuitive mental picture of what the free product of two groups represents. From what I understood, the free product $G\ast H$ is the group whose elements are the reduced words of the form $g_1 h_1 g_2 h_2 \dots g_k h_k$.
But let's take $G = (\mathbb{Z}, +)$ and $H = S_{4}$ the 4-th symmetric group. What I don't get is what $g_1 h_1 g_2 h_2 \dots g_k h_k$ would exactly mean in this case, since the operation in $G$ is the addition and the operation in $H$ is the composition of function. When we write $g_1 h_1$ what is the operation performed between $g_1$ and $h_1$? Addition or function composition? I am looking for an intuitive mental picture of the objects we are manipulating.
A word is a finite sequence, nothing more and nothing less.
The product operation on $G*H$ is done in two steps. The first step is concatenation of words: you are given two reduced words, each of which is a sequence, and you place the two sequences, one following the other, to make a longer sequence. So, for example, if you want to write the product of the reduced words $g_1h_1$ and $g_2h_2$ then your first step is to concatenate those two length 2 sequences to get a length 4 sequence $g_1h_1g_2h_2$.
The next step is reduction. The example that I just gave is already reduced, and so you are done: the product of the pair of length 2 reduced words $g_1h_1$ and $g_2h_2$ is the length 4 reduced word $g_1h_1g_2h_2$. The example you ask about is even simpler: the product of the two length $1$ reduced words $g_1$ and $h_1$ is the length $2$ word obtained by concatenation, namely $g_1h_1$; this word is, again, already reduced.
But there are other examples where the result of concatenation produces an unreduced word. Let's say you want to compute the product of the two reduced words $g_1 h_1 g_2$ and $g_3 h_3$. As said the first step is concatenation which produces the length 5 word $g_1h_1g_2g_3h_3$. However, this word is unreduced. To reduce it, you carry out an inductive procedure. For this example, you start with the subword $g_2g_3$ and you carry out that product using the given group operation on $G$ (the words you use to describe that group operation are irrelevant; you can call that operation addition, or multiplication, or whatever). Let's say that you do the computation and you come up with $g_2g_3 = g'$. If $g'$ is not the identity element of $G$ then you make a substitution $g_1h_1g'h_3$; so you have found a reduced word and the induction is done. If on the other hand $g'$ is the identity element of $G$ then you remove $g_2g_3$ to get a shorter unreduced word $g_1h_1h_3$. Now you repeat: calculate $h_1h_3$ using the given group operation on $H$, et cetera.
To summarize: the group operation on $G*H$ is concatenation followed by reduction. The place where the group operations of $G$ and $H$ come into play is in the reduction step; those operations have nothing to do with the concatenation step.