Answer by ℋolo for Distinction between Built-in and Definable Skolem Functions

$\newcommand{\L}{\mathcal{L}}$First let me define better those 2 terms:

$T_\L$ is a Skolem theory(or, has built in Skolem functions, same thing) if:

For every $n+1$-ary formula $φ$, there exists in$\L$ an $n$-ary function symbol $f$ such that $T\models \forall \overline x (\exists y \varphi(\overline x,y) \rightarrow \varphi (\overline x,f(\overline x)))$

That is, the Skolem function is a symbol in the language.

$T_\L$ has definable Skolem functions if:

For every $n+1$-ary formula $φ$, there exists $n+1$-ary functional relation $ψ$ such that $T\models \forall \overline x (\exists y \varphi(\overline x,y) \rightarrow ∀ z(\psi(\overline x,z)→\phi(\overline x,z)))$.

Now for the questions:

1. We distinguish between the 2 because the fact that there exists a functional relation that behave like the above, does not implies that there exists such symbol $f$. Given that $T_\L$ is Skolem theory, we can simply define $ψ(\overline x, z)=φ(\overline x,z)∧z=f(\overline x)$, I believe you can check that this is indeed a functional relation that satisfy the definition of definable Skolem function. On the other hand, given $ψ$, we can't say anything about the existence of $f$

2. No, take any theory without any function symbols, and you get a theory that is not Skolem theory.

Now, that being said, it is correct that we can just add function symbol for each formula $φ$ and add the axioms $x=f_{\varphi }(w) \iff \psi (w,u)$. The new theory conservative extension to the old theory. If the new theory definable Skolem functions again, we can repeat this processes, doing this recursively and we get that the union of all of those is a conservative extension of the original theory and is a Skolem theory.