In Definition 8.6 of Joy of Cats the authors write (I have changed the wording to avoid explaining the conventions that the authors introduce earlier which would lengthen the post unnecessarily),
Let $(\mathbf{A},U)$ be a concrete category over $\mathbf{X}$ (i.e, where $\mathbf{A}$ is a category and $U:\mathbf{A}\to\mathbf{X}$ is a faithful functor).
An $\mathbf{A}$-morphism $A\overset{f}{\to} B$ is called initial provided that for any $\mathbf{A}$-object $C$ and for the $\mathbf{X}$-morphism $U(C)\overset{g}{\to} U(A)$ there exists a unique $\mathbf{A}$-morphism $C\overset{g^\ast}{\to} A$ such that $U(g^\ast)=g$ whenever for the $\mathbf{X}$-morphism $U(C) \overset{U(f)\circ g}{\longrightarrow} U(B)$ there exists a unique $\mathbf{A}$-morphism $C\overset{h^\ast}{\to} B$ such that $U(h^\ast)=U(f)\circ g$.
An initial morphism $A\overset{f}{\to} B$ for which $U(f)$ is a monomorphism is called an embedding.
My questions are,
What is the reason for calling the for calling the $\mathbf{A}$-morphism initial in (1) of the above definition?
What would be the problem if we would simply define an embedding as follows,
2.$^\ast$ An morphism $A\overset{f}{\to} B$ for which $U(f)$ is a monomorphism is called an embedding.
I thought that an answer to (1) could be found by showing that the morphism $f$ is the initial object of some suitable category. But footnote 45 confused me where it is written that,
"Notwithstanding their names, the concepts of initial object and initial morphism are unrelated."
As asked for in the comments, here is what I've been thinking. I've been a bit too optimistic and this turns out to be more subtle than I initially thought.
As a slogan, I think we can say that
Of course, a terminal object of some category is an initial object of the opposite category, but this shouldn't count here if you ask me.
Let me fix a setup and some terminology for convenience: We have a faithful functor $|\cdot|\colon A\to X$ which allows us to consider $A$-morphisms as (certain) $X$-morphisms. That is, up to an isomorphism of categories, we can and will assume that $A(a,b)\subset X(|a|,|b|)$.
Let $f\colon a\to b$ be an $A$-morphism. It is called initial if and only if the following holds: for every $c\in A$, an $X$-morphism $|c|\to |a|$ is actually an $A$-morphism $c\to a$ (i.e., an element of the subset $A(c,a)\subset X(|c|,|a|)$) if and only if the composite $|c|\to |a|\to |b|$ is actually an $A$-morphism $c\to b$.
From here, I see two ways to go: Either, we may ask if $f\colon a\to b$ is in some sense initial or terminal in some arrow category, or (motivated by the initial topology), we may forget $a$ itself and only remember $b$, $|a|$ and the $X$-morphism $f\colon |a|\to |b|$, since the defining property is actually independent of $a$. I will concentrate on the latter part, mentioning the former idea only shortly at the very end.
To formulate the idea more precisely, let $b$ be an object of $A$, $x$ an object of $X$ and an $X$-morphism $f\colon x\to |b|$. (Think $x = |a|$.) Let me define a lift of $x$ compatible with $f\colon x\to |b|$ to consists of an object $c\in A$, together with an $X$-morphism $\varphi\colon |c|\to x$ such that the composite $f\circ \varphi\colon c\to b$ is an $A$-morphism. The $f$-compatible lifts of $x$ form a category, where a morphism $(c,\varphi)\to (c',\varphi')$ consists of an $A$-morphism $g\colon c\to c'$ such that $\varphi'\circ g = \varphi$.
Here is the upshot:
Proposition 1 – An $A$-morphism $f\colon a\to b$ is initial if and only if $a$, together with the identity map $\mathrm{id}\colon |a|\to |a|$, is terminal in the category of lifts of $|a|$ compatible with $f\colon |a|\to |b|$.
Proof: Note that $f\colon a\to b$ is initial if and only if $A(c,a) = \{g\in X(|c|,|a|)\mid f\circ g\in A(c,b)\}$ for every $c\in A$. On the other hand, consider the sub-functor $F\subset X(\cdot,|a|)\colon A^{op}\to(Set)$ mapping $c$ to $\{g\in X(|c|,|a|)\mid f\circ g\in A(c,b)\}$. It's category of elements is the opposite of the category of $f$-compatible lifts of $|a|$. Therefore, $f\colon a\to b$ is initial if and only if $A(\cdot,a) = F$, which is the case if and only if $(a,\mathrm{id}_{|a|})$ is initial in the category of elements of $F$, i.e., if and only if it is terminal in the category of lifts of $|a|$ compatible with $f$, as claimed.
Alternative, more elementary proof: Suppose that $f\colon a\to b$ is initial and let $(c,\varphi)$ be any lift of $|a|$ compatible with $f\colon |a|\to |b|$. That is, $\varphi\colon |c|\to |a|$ is such that the composite $f\circ \varphi\colon |c|\to |b|$ is actually an $A$-morphism $c\to b$. By assumption, this implies that $\varphi$ is actually an $A$-morphism $c\to a$ which is in fact the unique morphism $(c,\varphi)\to(a,\mathrm{id}_{|a|})$. Conversely, if $(a,\mathrm{id}_{|a|})$ is terminal, and if $h\colon |c|\to |a|$ is any morphism such that $f\circ h$ is an $A$-morphism $c\to b$, then $(c,h)$ is an $f$-compatible lift of $|a|$ and so there is a unique $A$-morphism $g\colon c\to a$ such that $\mathrm{id}_{|a|}\circ g = h$. Therefore, $g = h$ is an $A$-morphism, as desired.
Proposition 2 – An $A$-morphism $f\colon a\to b$ is initial if and only if the category of lifts of $|a|$ compatible with $f$ has a terminal object $(c,\varphi)$ such that $\varphi$ is an $X$-isomorphism.
Proof. Proposition 1 proves one implication. Conversely, suppose that $(c,\varphi)$ is terminal where $\varphi\colon |c|\to |a|$ is an $X$-isomorphism. Since $(a,\mathrm{id}_{|a|})$ is a lift of $|a|$ compatible with $f$, there is a unique $A$-morphism $g\colon a\to c$ such that $\varphi\circ g = \mathrm{id}_{|a|}$, so that $\varphi^{-1} = g$ is an $A$-morphism. Therefore, $\varphi$ is an isomorphism $c\cong a$ over $b$. Thus, $(a,\mathrm{id}_{|a|})$ is itself terminal in the category of $f$-compatible lifts of $|a|$ and the rest follows from Proposition 1.
It can also be shown:
Proposition 3 – Let $f\colon x\to |b|$ be an $X$-morphism where $b$ is an object of $A$. Suppose that $(c,\varphi)$ is a terminal lift of $x$ compatible with $f$, where $\varphi$ is a monomorphism. Then $f\circ \varphi\colon c\to b$ is an initial $A$-morphism.
Right now, I cannot come up with an example where $\varphi$ is not a monomorphism. (Anyone?) But one where it is not an isomorphism is easy to find. For example, let $X = (Set)$, $A = (Grp.)$ with the obvious forgetful functor, $b = (\mathbb Z,+)$ and $x = \{0,1\}$. Then $c = (\{0\},+)$ and $\varphi\colon |c| = \{0\}\to \{0,1\} = x$ is the inclusion and certainly not surjective. Nonetheless, the inclusion $c\to b$ is of course initial.
As promised, here is a short remark concerning terminality of $f$ itself instead of $a$: Trivially, $f$ is a weak terminal object in the full sub-category of the slice $A/b$ of those morphisms which factor through $f$. If $f$ is an initial morphism, then this category has another description, yet again relying only on $|a|$ and $f\colon |a|\to |b|$, yielding the following:
An initial morphism $f\colon a\to b$ is weakly terminal in the full sub-category of $A/b$ consisting of those morphisms $c\to b$ such that such that the underlying $X$-morphisms $|c|\to |b|$ factors through $f\colon |a|\to |b|$. But this doesn't seem to uniquely define $a$ or the initiality of $f$.