I am interested in the way the different generalized definitions of measurable functions relate to each other. Here are a few I am considering:
From Folland
If $(X, \mathcal{M})$ and $(Y, \mathcal{N})$ are measurable spaces, a mapping $f: X \to Y$ is called $(\mathcal{M}, \mathcal{N})$-measurable, or just measurable when $\mathcal{M}$ and $\mathcal{N}$ are understood, if $f^{-1}(E) \in \mathcal{M}$ for all $E \in \mathcal{N}$.
From Rudin
If $(X, \mathcal{M})$ is a measurable space, $(Y, \tau)$ is a topological space, and $f$ is a mapping of $X$ into $Y,$ then $f$ is said to be measurable provided that $f^{-1}(V) \in \mathcal{M}$ for every $V \in \tau.$
Finally from my first- year graduate course in analysis, in which we define measurable spaces and measurability in terms of $\sigma$-rings as opposed to $\sigma$-algebras, we have this proposition.
Let $(X, \mathcal{S})$ be a measurable space, $B$ a Banach space, and $f$ a function from $X$ to $B.$ Then $f$ is $\mathcal{S}$-measurable if and only if
1. $f(X)$ is a separable subset of $B,$ and
2. $f^{-1}(U) \cap C(f) \in \mathcal{S}$ for every open ball, $U,$ in $B$
where $C(f) = \{ x \in X : f(x) \neq 0 \}.$
The proposition above seems almost equivalent to Rudin's definition if we restrict it to $\sigma$-algebras since $$f^{-1}(U) \cap C(f) = f^{-1}(U) \cap f^{-1}(B \setminus \{0\}) = f^{-1}(U \setminus \{0\})$$ and singletons are closed in $B.$ But does the requirement that $f(X)$ be separable change the class of measurable functions if we let $(Y, \tau) = B$ with its topology induced by the norm?
Furthermore, Rudin's definition agrees with Folland's when $\mathcal{N}$ is the collection of Borel subsets, and we can always generate a $\sigma$-algebra from a topology. However, are there $\sigma$-algebras that cannot be generated by a topology? If so then it would seem that Folland's definition is "more general", otherwise they would be equivalent wouldn't they?
(In my opinion) Folland's definition is the "real" one. You are correct that Rudin's definition is secretly a special case of Folland's, if you put the Borel $\sigma$-algebra on $Y$ (that is, the fact that the inverse image of every open set is measurable implies that the inverse image of every Borel set is measurable).
Your third quote, with $\mathcal{S}$-measurability, looks closer to what I would call "strong measurability", which arises in the theory of vector-valued integration. See also: https://en.wikipedia.org/wiki/Bochner_measurable_function