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.
Godel produces a sentence $\varphi$. What Godel proves is that - assuming $PA$ is consistent - $\varphi$ is true but not provable in $PA$. (I'm assuming the theory we're looking at is "PA," here - but we can of course replace $PA$ with any sufficiently strong recursively axiomatized theory, such as PM, ZFC, NF, ...)
This proof goes through perfectly inside the theory $PA$. There's no contradiction, though, because - in order to conclude that $\varphi$ is true - $PA$ would have to know that $PA$ is consistent. So, instead of a paradox, we get Godel's second incompleteness theorem: that, if $PA$ is consistent, $PA$ doesn't prove "$PA$ is consistent."
I'm being ahistorical here - in fact, what Godel proved was slightly weaker, and Rosser was the one who brought the hypothesis down to "$PA$ is consistent" - but this is the meat of the situation.
Best Answer
As your quotes from Berlinski suggest, the heart of Godel's statements is arguably not in the unprovability itself, but in the fact that the system is able to 'talk about itself' - that is, that statements about provability can be cast as strictly arithmetic questions. It might help to familiarize yourself with another undecidability result which goes through roughly the same route: Matiyasevich's Theorem about the solvability of diophantine equations (i.e., the question 'does this polynomial in $x$, $y$, $z$, $w$, etc. ever evaluate to zero at some integer values of $x$, $y$, $\ldots$?'). The root of the proof lies in showing that diophantine equations are expressive enough to represent all recursively enumerable sets; what Godel does is essentially the same, showing that Peano Arithmetic (or any similar system) is expressive enough that questions about 'provability' can be cast as arithmetic questions.
As for why unprovable statements need admission, the point isn't simply that they're unprovable but that they're unprovable and true - that is, that the set of 'things we can prove' will always be a proper subset of 'things that are true'. Obviously this isn't a crippling discovery; mathematics didn't stop as of Godel! But it is arguably a profound one, and one that has substantial practical implications: for instance, as noted above there can't possibly be an algorithm for solving diophantine equations; there can't be algorithms for finding a 'minimal forbidden minor' categorization of many sets of graphs; there can't be algorithms for solving the Word Problem for groups; etc, etc.