are quotients by equivalence relations "better" than surjections?
Are inclusions in some way better than arbitrary injections? Are maps to sets of equivalence classes somehow better than arbitrary quotients?
This is a simple-minded answer, but I would say yes because there is a proper class of distinct injections into $Y$ --- the injecting set $X$ can live anywhere in the set-theoretic universe --- whereas inclusions into $Y$ are effectively just subsets of $Y$. Under the natural notion of equivalence for injections, distinct subsets give inequivalent injections, so you could say that going from injections to inclusions is a matter of selecting one distinguished, canonical element from each equivalence class of injections.
I would say that the injections into $Y$ are classified by the subsets of $Y$, and similarly the quotients of $X$ are classified by the equivalence relations on $X$.
It seems that you've got factorization of maps covered, so let me address the question of why canonical quotient maps and canonical inclusions are "better".
Given a set $X$, in general there is a proper class of injections $Y \to X$. However, many of these are isomorphic, where injections $i : Y \to X$ and $j : Z \to X$ are isomorphic if there is an isomorphism $k : Y \to Z$ such that $i = j \circ k$. The isomorphism classes of injections into $X$ are the subobjects of $X$. In fact, there are only set-many subobjects of $X$ (in category-theoretic language, sets form a well-powered category). It is pesky to work with set-many proper equivalence classes, so we instead look for a set $P(X)$ of injections into $X$, one from each isomorphism class. We may additionally require some nice properties, for instance, if $i : X \to Y$ is in $P(Y)$ and $j : Y \to Z$ is in $P(Y)$, we would expect $j \circ i : X \to Z$ to be in $P(Z)$. One can come up with a wish list of such nice closure conditions, here's another one: if $i : Y \to X$ and $j : Z \to X$ are in $P(X)$, and there is (a unique) $k : X \to Z$ such that $i = j \circ k$, then $k$ is in $P(Z)$.
We know the answer, of course, just take $P(X)$ to be the canonical subset inclusions into $X$. This is not the only choice of such representative inclusions, but it's a pretty good one.
We may therefore say that the canonical inclusions of subsets are "better" because they are the canonical representatives of subobjects (equivalence classes of injections).
The answer for quotient maps and surjections is dual. Consider equivalence classes of surjections, quotiented by isomorphism. There are only set-many such classes, therefore sets form a well-copowered category. (Some people say "cowell-powered" but then why not call it "ill-powered"?) This time we look for a set $Q(X)$ of surjections from $X$, each representing one equivalence class of surjections from $X$. We may take $Q(X)$ to be the set of all canonical quotient maps $X \to X/{\sim}$, or just the set of all equivalence relations on $X$. Once again, canonical quotient maps are "better" because they are the distinguished representatives of isomorphism classes of surjections.
The nLab page you're looking for is called factorization systems. Here is my favorite one, which I think answers your question. In any category with finite limits and colimits, every morphism $f : X \to Y$ has a canonical factorization
$$X \to \text{coim}(f) \to \text{im}(f) \to Y$$
where $\text{im}(f)$, the regular image, is the equalizer of the cokernel pair of $f$ (this is the "nonabelian" version of "kernel of the cokernel") and $\text{coim}(f)$, the regular coimage, is the coequalizer of the kernel pair of $f$ (again, the "nonabelian" version of "cokernel of the kernel"). These two constructions are categorically dual and so, among other things, the coimage-image factorization of $f$ in the opposite category is the same sequence of maps but in the opposite order.
In $\text{Set}$, the coimage and image are both the image of a function in the usual sense, but computed in different ways, which I think match the distinction you're getting at. $\text{coim}(f)$ is computed, more or less, by constructing the equivalence relation on $X$ defined by $x_1 \sim x_2 \Leftrightarrow f(x_1) = f(x_2)$, then quotienting $X$ by it. $\text{im}(f)$ is computed in a categorically dual way, although it looks a little strange at first: by first constructing the pushout $Y \sqcup_X Y$, then isolating the subset of $Y$ of elements which are sent to the same element by both of the canonical maps $Y \to Y \sqcup_X Y$.
In particular, the factorization you want for an injection is the regular image factorization, and the factorization you want for a surjection is the regular coimage factorization, so they are in fact categorically dual. The full coimage-image factorization combines these.
It's a nontrivial theorem that the map $\text{coim}(f) \to \text{im}(f)$ is an isomorphism in $\text{Set}$. It's also an isomorphism in any abelian category and in $\text{Grp}$ (this is an abstract form of the first isomorphism theorem), but in general it's just both a monomorphism and an epimorphism. A very instructive example is $\text{Top}$, where $\text{coim}(f)$ is the set-theoretic image topologized as a quotient of $X$, and $\text{im}(f)$ is the set-theoretic image topologized as a subspace of $Y$. (Note that these coincide for compact Hausdorff spaces!)