Here is the statement and proof in question.
I don't understand how they can say that $W$ has a quotient topolgy from $X\times Z$ and that there is a quotient map $g$. It seems like this what we are trying to prove. To me, a quotient map is a certain kind of surjective, continuous map. I also don't understand why we want to show that $h$ is continuous.
Also if someone could tell me what the subsript $c$ means here that would be greatly appreciated:
I think Hatcher here means as quotient topology the most rich topology making the map in question continuous. In our situation he defines $W=Y \times Z$ as set and since he wants $f\times 1$ to be a quotient map he finest topology on $W$ making this function continuous, i.e. we declare a subset $A \subset W$ to be open iff $(f \times 1)^{-1}(A)$ is open in $X \times Z$. In other terms we are putting on $W$ the final topology with respect to the map $f \times 1$. See the wikipedia article for the definition https://en.wikipedia.org/wiki/Final_topology This is exactly the topology you are considering for quotient maps, i.e. maps whose underlying function of set is a function from a set $T$ to its quotient $T/\sim$ with respect to $\sim$ an equivalence relation.
Now we want to prove that the product topology on $Y \times Z$ and this quotient topology coincide, if you spell the condition explicitly you will see that this is equivalent to proving that the identity function $h \colon Y \times Z \rightarrow W$ is continuous.
For your second questions, usually the notation $X_c$ refers to the compactly generated topology associated to the space $X$. A good reference is https://ncatlab.org/nlab/show/compactly+generated+topological+space basically we want to modify the starting topology on $X$ in such a way that all its closed subsets $C \subset X$ will be compactly closed, that is for every continuous function $f \colon K \rightarrow X$ from $K$ compact space $f^{-1}(U)$ must be closed.
A famous example of class of compactly generated space is the class of CW complexes.
The point to bring this up in the proof I think is the following: in general the product of two compactly generated spaces $X$, $Y$ is not compactly generated, thus you consider $(X \times Y)_c$ to have again a compactly generated space.
For example if $X$ and $Y$ are two CW complexes $X \times Y$ needs not to be a CW complex, while $(X \times Y)_c$ admits the structure of CW complex.