Is it possible to deduce Godel's first incompleteness theorem from Chaitin's incompleteness theorem?
Gödel's incompleteness theorem, in its modern form using Rosser's trick, only requires that (beyond being effective and sufficiently strong), the theory must be consistent. There is no requirement that the theory has to be $\omega$-consistent or meet any soundness assumption beyond simple consistency. You cannot apply Chaitin's theorem in its usual form to these theories, in general, because the usual form of Chaitin's theorem assumes more (e.g. many proofs of Chaitin's theorem assume as an extra soundness assumption that the theory only proves true statements.)
Many of the "alternate proofs" also require stronger assumptions than the standard proof of the incompleteness theorem. You have to be very careful when reading about this to check which assumptions are included in each theorem.
In the particular proof of Chaitin's theorem presented by Kritchman and Raz, however, they do not need to assume any particular soundness beyond just consistency. I am going to explain this in detail.
They do need to assume that $T$ is sufficiently strong. In particular, they assume that if $n$ and $L$ satisfy $K(n) < L$, then $T \vdash \hat K(\dot n) < \dot L$. Here $\dot n$ and $\dot L$ are terms of the form $1 + 1 + \cdots + 1$ corresponding to $n$ and $L$, and $\hat K$ is a formula of arithmetic defined directly from the definition of $K$. (Note that the set of pairs $(n,L)$ with $K(n) < L$ is recursively enumerable, so there is no real issue in assuming $T$ proves all true sentences of that form.)
Given the assumption on $T$, their proof goes as follows (rephrased in more precise terms):
Begin proof
Given $L$, we can make a program $e_L$ that does the following:
Search for any $n$ such that $T \vdash \hat K (\dot n) > \dot L$. We do this by searching through all $T$-proofs in an exhaustive manner.
Output the first such $n$ we find, if we ever find one.
Because we can code $L$ into the program as a fixed constant, using the standard coding methods, the length $|e_L|$ of $e_L$ no worse than $2\log(L) + C$ for some constant $C$. In particular, we can fix an $L$ with $|e_L| < L$. Assume such an $L$ is fixed.
For this $L$, suppose $e_L$ returns a value, $n$. Then $K(n) \leq |e_L| < L$. By our assumption that $T$ is sufficiently strong, this means $T$ proves $\hat K (\dot n) < \dot L$.
But if $e_L$ returns $n$ then $T$ also proves $\hat K(\dot n) > \dot L$. This means that if $e_L$ returns a value then $T$ is inconsistent. So, if we assume $T$ is consistent, then $e_L$ cannot return a value. This means that, for our fixed $L= L_T$, $T$ cannot prove $\hat K(\dot n) > \dot L$ for any $n$.
Thus, if we take $n = n_T$ to be such that $K(n) > L$, we have that $\hat K (\dot n) > \dot L$ is a true statement that is not provable in $T$.
End proof
This proof just given (in italics) proves:
If $T$ is a consistent, effective theory of arithmetic that proves every true formula of the form $\hat K(\dot n) < \dot L$, then there is a true statement of the form $\hat K(\dot n) > \dot L$ that is not provable in $T$.
This is almost the same as Gödel's incompleteness theorem. The only difference is that the usual proof of the incompleteness theorem gives us an explicit formula that it is true but not provable in $T$ (namely, the Rosser sentence of $T$). On the other hand, the proof in italics requires us to find $n_T$ in order to have an explicit example, and there is no algorithm to do this.
This is one motivation for what I think of as the "standard form" of Chaitin's theorem. In that form, we look instead for unprovable sentences of the form ($\dagger$): $(\exists x)[\hat K(x) > \dot L]$.
Because we can compute $L= L_T$, we can compute a specific sentence of that form for $T$. But, for the proof to work, we need to have an actual $n$ such that $T \vdash \hat K(\dot n) > \dot L$. So we have to add an additional soundness assumption to the theorem, namely that if $T$ proves a sentence of the form ($\dagger$) then there is an $n$ such that $T$ proves $\hat K(\dot n) > \dot L$.
Overall, in this version of Chaitin's theorem, we have an explicit sentence, but with a stronger soundness assumption. The proof of Gödel's incompleteness theorem using Rosser's method gives us an explicit sentence without any stronger soundness assumption.
As others have said, the very distinction you're trying to draw is fuzzy to the point of meaninglessness, and there are very natural examples of undecidable (with respect to any of the usual theories) sentences. However, there's a very nice fact which hasn't been mentioned yet and I think puts the final nail in the coffin:
For every "reasonable" theory $T$, there is a Diophantine equation $\mathcal{E}_T$ such that $\mathcal{E}_T$ has no solutions but $T$ cannot prove that $\mathcal{E}_T$ has no solutions.
It's hard to get more concrete and arithmetic than Diophantine equations!
Here as usual "reasonable" means "computably axiomatizable, consistent, and extending $Q$" (although even that is overkill: we can replace "extending $Q$" with "interpreting $R$" or even weaker hypotheses - see here, Section $4$). The result above is a more-or-less immediate consequence of the proof of the MRDP theorem: basically, we can assign to $T$ a Diophantine equation $\mathcal{E}_T$ such that putative solutions to $\mathcal{E}_T$ correspond to putative contradictions in $T$. Of course, the $\mathcal{E}_T$s so produced are truly god-awful, but they are honest-to-goodness Diophantine equations
Best Answer
To the extent that our "axioms" are attempting to describe something real, yes, axioms are (usually) independent, so you can't prove one from the others. If you consider them "true," then they are true but unprovable if you remove the axiom from the system. In that sense, the smaller system has "true" but unprovable theorems.
But the "trueness" of Gödel's statement is a bit more complicated.
Let's say it turned out that the Goldbach conjecture was undecidable. To me, that would mean that it is "intuitively true," since if it was false, we could find a counter-example that was a finite statement. The fact that we can't provide a counter-example, however, is not enough to prove it is true. This might seem strange, even absurd.
One way I like to think of it is that proofs are finite things, but we are often trying to prove things about infinitely many numbers. Induction, for example, can be thought of as a finite way of outlining an infinite proof.
Intuitively, what Gödel showed is that (under enough complexity) there are always theorems that have infinite proofs but for which there are no finite proofs. For example, Goldbach might be one of those cases. Each $2n, n\geq 3$ might be expressible as the sum of two primes, but if we can't outline that proof finitely, we are stuck.
A statement like Gödel's - rougly, "This theorem does not have a (finite) proof from these axioms" - is such an example. We can enumerate all finite logical proofs, and check if it proves our theorem. This would yield an infinite proof of the result. So in that sense, it is "true." But it obviously can't be proven.
The fact that we can't resolve this statement from these axioms means, intuitively, that it is true. Our intuition about the natural numbers says this ought to be true, and that our axioms didn't fully capture our intuition.