Does Gödel's second incompleteness theorem provide a specific sentence for his first theorem? Ebbinghaus' Mathematical Logic  on p184

7.8 Gödel's First Incompleteness Theorem.  Let $\Phi$ be  consistent and
R-decidable  and  suppose  $\Phi$  allows  representations.  Then  there  is  an  $S_{ar}$-
sentence $\phi$  such that neither $\Phi \vdash \phi$  nor  $\Phi \vdash \neg \phi$.

p185

7.10 Gödel's Second Incompleteness Theorem.  Let $\Phi$  be consistent  and R-decidable  with $\Phi \supset \Phi_{PA}$.  Then  $$ \text{not} \quad \Phi  \vdash Consis_\Phi.$$

On the relation between the two theorems, p184 says

A refinement of the above argumentations (in Godel's first Incompleteness
Theorem) leads to results concerning the
consistency of mathematics. In particular, Godel's Second Incompleteness
Theorem, which we shall now derive, shows that the consistency of a suffi-
ciently rich system cannot be proved using only the means available within
the system.

and https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems#Second_incompleteness_theorem says

This theorem (Godel's second incompleteness theorem) is stronger than the first incompleteness theorem because the statement constructed in the first incompleteness theorem does not directly express the consistency of the system. The proof of the second incompleteness theorem is obtained by formalizing the proof of the first incompleteness theorem within the system F itself.

I have trouble understanding the relation between the two theorems in the above sources.
Does the first theorem roughly say that if a set of sentences can represent all the computable functions, the deductive theory derived from the set can't have both proof consistency and proof completeness: either one of them, or neither?
Does the second theorem  say almost the same as the first theorem, except that while the first theorem does not give a specific sentence which can be neither proved nor disproved from the set, the second theorem provides $Consis_\Phi$ as such a sentence?
Does the second theorem imply the first?
 A: 
Does the first theorem roughly say that if a set of sentences can represent all the computable functions, the deductive theory derived from the set can't have both proof consistency and proof completeness: either one of them, or neither?

Sure, that's one way to think about it (though you left out the crucial assumption that the set of sentences should also be decidable).

Does the second theorem  say almost the same as the first theorem, except that while the first theorem does not give a specific sentence which can be neither proved nor disproved from the set, the second theorem provides $Consis_\Phi$ as such a sentence?

Not quite.  The second theorem provides $Consis_\Phi$ as a sentence that cannot be proved from $\Phi$.  However, it's possible that $Consis_\Phi$ could be disproved from $\Phi$.  Intuitively, this means that "$\Phi$ proves its own inconsistency".  You might think that this means $\Phi$ must be inconsistent (contrary to the assumption that it was consistent), but that's not necessarily the case--the sentences that $\Phi$ proves don't necessarily have to be true (when interpreted in the natural numbers).  For instance, if you start with a set of sentences $\Phi$ and then let $\Phi'=\Phi\cup\{\neg Consis_\Phi\}$, then $\Phi'$ proves $\neg Consis_\Phi$ and thus also $\neg Consis_{\Phi'}$.  But $\Phi'$ is still consistent, exactly because $\Phi$ did not prove $Consis_\Phi$, so adding its negation to the theory does not lead to a contradiction.

Does the second theorem imply the first?

No, as explained above.  The second theorem implies a weaker version of the first where you add an additional assumption that implies $\Phi$ does not prove its own inconsistency.  For instance, you could assume that every sentence in $\Phi$ is actually true when interpreted in the natural numbers, and so $\Phi$ cannot prove any false statement about the natural numbers.  In particular, since $\neg Consis_\Phi$ is false (i.e., $\Phi$ is not actually inconsistent), this means $\Phi\not\vdash \neg Consis_\Phi$.
