The following question is taken from $\textit{Categories}$ by Horst Schubert.
Let $C$ be a category, for each ordered pair of objects $(A,B)$ of $C$ a (possibly empty) set $[A,B]_C$ called the set of $\textit{morphisms}$ from $A$ to $B$. If it is obvious from the context which category is meant, $[A,B]_C$ will be replaced by $[A,B].$
Let $C$ be a category with a zero object $0$. if $A$ and $B$ are any objects, then there is exactly one morphism $A\rightarrow B$ which factors through $0$; i.e., it can be represented in the form $A\rightarrow 0\rightarrow B$. We call it the $0-$ morphism and it is usual to denote it by $0$ (better $0_{A,B}$). $0:A\rightarrow B$ does not depend on the choice of a zero object $0$ in $C$. If $0'$ is another zero object, consider $A\rightarrow 0\rightarrow 0'\rightarrow B.$
Questions: What do the following two maps mean, or rather what are they suppose to equal to?
$A \xrightarrow{\normalsize\textit{f}} 0\xrightarrow{\normalsize\textit{g}}B,$ $g\circ f=?$
$A \xrightarrow{\normalsize\textit{f'}} 0\xrightarrow{\normalsize\textit{g'}}0'\xrightarrow{\normalsize\textit{h'}}B,$ $h'\circ g'\circ f'=?$ (I assigned letters to the maps since it might be easier to refer to them individually.)
Also, how does one get from $0:A\rightarrow B$ to the compositional maps in the definition above?. Lastly, what does it when a morphism factors through $0$?
Thank you in advance
By definition, to say that $0$ is a zero object means that for any object $A$ in your category, there is a unique morphism $f_A:0\rightarrow A$ and a unique morphism $g_A:A\rightarrow 0$.
Now, suppose $A$ and $B$ are objects in your category. If we select some zero object $0$, then we obtain a special morphism from $A$ to $B$: the morphism $f_B\circ g_A$.
But what if we select another zero object $\hat{0}$? Then we obtain another special morphism from $A$ to $B$: $\hat{f}_B\circ \hat{g}_A$.
Proposition: The two special morphisms from $A$ to $B$ are, in fact, the same. That is, we have an equality $f_B\circ g_A = \hat{f}_B\circ \hat{g}_A$.
Proof: Since $0$ is initial, we have a unique morphism $f_{\hat{0}}:0\rightarrow \hat{0}$. The composition $\hat{f}_B\circ f_{\hat{0}}:0\rightarrow B$ is a morphism from $0$ to $B$. Since there is a unique such morphism, we have $f_B = \hat{f}_B\circ f_{\hat{0}}.$
Likewise, we have a morphism $\hat{f}_0: \hat{0}\rightarrow 0$. We similarly find that $g_A = \hat{f}_0 \circ \hat{g}_A$.
Thus, we have $f_B\circ g_A = \hat{f}_B\circ f_{\hat{0}} \circ \hat{f}_0\circ \hat{g}_A$.
But what is $f_{\hat{0}}\circ \hat{f}_0$? Well, it is a morphism from $\hat{0}$ to $\hat{0}$. Since $\hat{0}$ is a zero object, there is only one morphism from $\hat{0}$ to itself. On the other hand, in any category, any object has an identity morphism from itself to itself. Thus, by uniqueness of morphism $f_{\hat{0}}\circ \hat{f}_0 = 1_{\hat{0}}$, where $1$ denotes the identity morphism.
Putting this all together, we have $$f_B\circ g_A = \hat{f}_B\circ f_{\hat{0}} \circ \hat{f}_0\circ \hat{g}_A = \hat{f}_B\circ 1_{\hat{0}}\circ \hat{g}_A = \hat{f}_B\circ \hat{g}_A.$$
Thus, the proof is completed $\square$
Based on this proposition, we may unambiguously talk about "the" special morphism from $A$ to $B$: simply select any zero object, and then form the composition $f_B\circ g_A$. Your book simply uses the notation $0_{A,B}:A\rightarrow B$ as the name of this composition.
As far as answering what the composition $f_B\circ g_A$ is equal to, that depends on the category. For example, in the category of groups or in the category of rings, $0_{A,B}$ is the constant function which sends every element of $A$ to the (additive) identity of $B$. In the category of pointed topological spaces, $0_{A,B}$ is the constant function which sends every element of $A$ to the "special" point of $B$.
But also keep in mind that objects in a category don't need to have elements, and the morphisms don't need to be functions in the usual sense. So, in general, answering what $0_{A,B}$ is, the best answer one can give is the recipe above: $0_{A,B}$ is the composition $f_B\circ g_A$ where $0$ is any zero object, and where $f_B:0\rightarrow B$ and $g_A:A\rightarrow 0$ are the unique morphisms one obtains from the definition of zero object.
I also wanted to address your last comment: what does it mean for a morphism to factor through something. Here is the official definition: Suppose $A,B,$ and $C$ are objects of some category and $h:A\rightarrow B$ is a morphism. Then $h$ is said to factor through C if there are morphisms $g:A\rightarrow C$ and $f:C\rightarrow B$ with the property that $h = f\circ g$.
Thus, $0_{A,B}$ factors through $0$ by definition (taking $f = f_B$ and $g= g_A$).