Let $\mathsf{PA_2}$ be a standard recursively axiomatized second-order Peano Arithmetic (e.g. the theory which Hilbert and Bernays call $\mathsf{Z}_2$). And interpret the wffs of $\mathsf{PA_2}$'s language with the 'full' semantics, where the second-order quantifiers are constrained to run over the full collection of sets of numbers.
Then (i) Gödel's theorem applies, of course, to $\mathsf{PA_2}$, so (assuming consistency) there will be a true canonical Gödel sentence $\mathsf{G}$ such that $\mathsf{PA_2} \nvdash \mathsf{G}$.
However, (ii) $\mathsf{PA_2}$, with the full semantics, is categorical by Dedekind's argument, so has only one model (up to isomorphism). In that one model, $\mathsf{G}$ is true, so $\mathsf{PA_2} \vDash \mathsf{G}$.
This shows that syntactic provability $\vdash$ falls short of semantic entailment $\vDash$ in the scecond-order case (there is no complete deductive system for second-order consequence, assuming the full semantics).
In summary, then, while the claim "$\mathsf{PA}$ settles all arithmetical truths" is false however we interpret it (where $\mathsf{PA}$ is first-order Peano Arithmetic), the situation with the corresponding claim "$\mathsf{PA_{2}}$ settles all arithmetical truths" is more complex. It depends whether you mean "semantically entails" or "deductively implies".
For vividness, it may well help to put that symbolically. We'll use $\{\mathsf{PA}, \vdash\}$ to denote the set of theorems that follow in $\mathsf{PA}$'s formal proof system, and $\{\mathsf{PA}, \vDash\}$ to mean the set of sentences semantically entailed by $\mathsf{PA}$'s axioms (given the standard semantics for first-order arithmetic). Similarly, we'll use $\{\mathsf{PA_{2}}, \vdash_2\}$ to mean the set of theorems that follow in $\mathsf{PA_{2}}$'s formal proof system, and $\{\mathsf{PA_{2}}, \vDash_{2}\}$ to mean the set of sentences semantically entailed by $\mathsf{PA_{2}}$'s axioms (given the "full" semantics for the quantifiers). Finally we'll use $\mathcal{T}_{A}$ to denote the set of truths of first order arithmetic (often called, simply, True Arithmetic); and we'll now use $\mathcal{T}_{2A}$, the set of truths of the language of second-order arithmetic (True Second-Order Arithmetic). Then we can very perspicuously display the relations between these sets as follows, using `$\subset$' to indicate strict containment:
$\{\mathsf{PA}, \vdash\} \;=\; \{\mathsf{PA}, \vDash\} \;\subset\; \mathcal{T}_{A}$
$\{\mathsf{PA_{2}}, \vdash_2\} \;\subset\; \{\mathsf{PA_{2}}, \vDash_{2}\} \;=\; \mathcal{T}_{2A}$.
The completeness of first-order logic which yields $\{\mathsf{PA}, \vdash\} = \{\mathsf{PA}, \vDash\}$ is, though so familiar, a remarkable mathematical fact which is enormously useful in understanding $\mathsf{PA}$ and its models. This has led some people to think the strict containment $\{\mathsf{PA_{2}}, \vdash_2\} \subset \{\mathsf{PA_{2}}, \vDash_{2}\}$ is a sufficient reason to be unhappy with second-order logic. Others think that the categoricity of theories like second-order arithmetic, and the second-order definability of key mathematical notions like finiteness, suffices to make second-order logic the natural logic for mathematics. Stewart Shapiro's Foundations without Foundationalism is a classic defence of the second position.
The issue here is discussed at some length in Ch. 29 (or Ch. 22 of the 1st edition) of my Gödel book, if you want more!
No. Compactness merely guarantees that $\mathsf{PA}$ (as is now standard I'll write simply "$\mathsf{PA}$" for first-order Peano arithmetic) has multiple models up to isomorphism. It does not prevent such models from all looking the same from the perspective of their internal first-order theories - that is, from being elementarily equivalent. Indeed, compactness applies to all first-order theories, including the complete ones. True arithmetic $\mathsf{TA}$ for example has $2^{\aleph_0}$-many non-isomorphic countable models, but is by definition complete.
The incompleteness of $\mathsf{PA}$ is a much stronger result, and does not follow from coarse considerations alone. Compactness shows that $\mathsf{PA}$ is not categorical (and indeed no first-order theory with infinite models is categorical). But even that is limited: it takes more than mere compactness to even show that $\mathsf{PA}$ isn't $\aleph_0$-categorical (we have to bring in Lowenheim-Skolem, or use Godel).
Best Answer
Your use of the word "intuitive" means that we're entering philosophical waters.
In ZFC, as you know, one can prove the formal version of the assertion "there is, up to isomorphism, only one model of the second-order Peano axioms". But as you are also aware, there are non-standard models of ZFC (assuming it's consistent), and these can have non-standard omegas.
Now to someone who adopts a certain kind of formalist attitude, there's no more to be said. For an intuitionist like Brouwer, on the other hand, our intuition of the natural numbers is basic, and not something that one can (or should even try) to formalize.
For most Platonists, it's simply assumed (an "article of faith") that the entities of mathematics exist in some "universe of abstractions". $\mathbb{N}$ is then a denizen of this universe. Our axiom systems, like PA and ZFC, capture some of the "facts" about this universe that we somehow know.
If you're asking for a precise mathematical definition of the natural numbers, I would say either "there isn't any", or "omega, working inside the ZFC axioms, or something along those lines".
Consider why you're sure that Goodstein's theorem holds for "the standard natural numbers". Surely it's because you believe in ZFC, in some sense. Or at least some fragment of ZFC. In other words, the power set of $\omega$ is ascribed some sort of reality.
Now consider ZFI, which is ZFC plus "there is an inaccessible cardinal". The consistency of ZFI does not follow from the consistency of ZFC. The statement Con(ZFI) is, however, a statement in PA (but of course not provable in PA). Are you sure it's true? Likewise for even stronger large cardinal axioms, or other extensions of ZFC that cannot be shown to be relatively consistent.
As it happens, John Baez and I carried on a lengthy discussion of these matters. Baez said:
Along the way we considered the "ultra-finitists", who consider even such a notion as $2^{100}$ as unclear.
You'll find the conversation here. Post 5 and post 6 especially focus on your question.
One last comment: regardless of one's philosophical stance, if someone says "This is true of the standard natural numbers", the purely mathematical content of this claim usually amounts to "this can be proved for $\omega$ in ZFC" (or some fragment of ZFC). At least that's how I would understand their assertion.