Yes, surjectiveness is a property of functions, and categories don't have to have anything to do with functions. There's a notion of a concrete category that does a reasonable job of capturing the notion of category where the arrows can be thought of as functions. Any category of algebraic objects (rings, groups, lattices, etc.) is a concrete category.
I would assume it's mentioned because the "dual-ish" statement that an injective homomorphism (in a category of algebraic objects) is precisely a monomorphism is true. (According to Wikipedia, sometimes "epimorphism" is defined to be "surjective homomorphism" by people outside of category theory, so it's important to show that the categorist's notion is not that.) Proof: In a category of algebraic objects, the forgetful functor $U$ is a right adjoint. $f : A \to B$ is a monomorphism if and only if the following diagram on the left is a pullback:$$
\require{AMScd}
\begin{CD}
A @= A @. @.@. @. @. UA @= UA \\
@| @VVfV @. @. @. @. @| @VVUfV \\
A @>>f> B @. @. @. @. @. UA @>>Uf> UB
\end{CD}$$
But then the right hand diagram is a pullback and so $Uf$ is injective. That proves that every monomorphism is an injective homomorphism. To show that every injective homomorphism gives rise to a monomorphism, assume $Uf$ is injective (i.e. a monomorphism in $\mathbf{Set}$). We can form the pullback of $f$ along itself as algebraic categories are complete producing the left hand diagram:$$
\begin{CD}
A\times_B A @>p>> A @. @.@. @. @. U(A\times_B A) @>Up>> UA \\
@VqVV @VVfV @. @. @. @. @VUqVV @VVUfV \\
A @>>f> B @. @. @. @. @. UA @>>Uf> UB
\end{CD}$$
The right hand diagram is again a pullback, but $(UA,id,id)$ is also a pullback of $Uf$ along itself so we have $Up : U(A\times_B A)\cong UA$ canonically, and similarly for $Uq$ so $Up = Uq$. Via faithfulness $p = q$ and $f$ is a mono follows. $\square$
All we really needed above was that $U$ is faithful and preserves kernel pairs and that the source category has them. $U$ usually won't preserve cokernel pairs. For example, the injections of a coproduct of an object with itself are a cokernel pair, but the underlying set of the coproduct of algebraic objects is usually not the disjoint union of the underlying sets of the objects. So dualizing the above proof does not work, even for only one direction, though this does not mean another proof couldn't be used.
As for right inverses, every epimorphism having a right inverse is equivalent to the axiom of choice in $\mathbf{Set}$. So even in $\mathbf{Set}$ having a right inverse is already a property one may find overly strong as a definition of epimorphism. If we consider the dual statement, "all monomorphisms split in $\mathbf{Set}$," it is almost but not quite true. The counterexample is any function from the initial object (i.e. the empty set) into any non-empty set. The point being there is less of a chance that "has a right inverse" will be a property one expects an epimorphism to have, but there's a good possibility one might expect "epimorphism" implies surjective homomorphism (in concrete categories) because that literally is the definition of "epimorphism" sometimes, and it often is true.
