Your question seems to boil down to *(after fixing an error)* the following:

> Any model $\mathfrak{M}$ of ACA$_0$ has a first-order part $Num(\mathfrak{M})$, which satisfies PA; why doesn't this mean that ACA$_0$ proves "PA has a model" (and hence its own consistency)?

Here ACA$_0$ is a conservative extension of the type Jech mentions (what he calls "$\Gamma$"): it can talk about models and prove basic model-theoretic results *(e.g. the soundness and completeness theorems)*, and is PA-provably a conservative extension of PA.

The issue is the following. Fix a model $\mathfrak{M}$ of ACA$_0$. We can indeed talk about $Num(\mathfrak{M})$ as a structure in $\mathfrak{M}$, and have the following:

> $(*)\quad$ Any set of sentences true in $Num(\mathfrak{M})$ is consistent.

However, we are *not* guaranteed that our model $\mathfrak{M}$ thinks that its first-order part actually satisfies PA. That is, the "obvious truth" of the PA axioms is not actually that obvious. 

This is an example of a failure of the $\omega$-rule: while for each axiom $\varphi$ of PA we do in fact have that "$Num(\mathfrak{M})\models \varphi$" (appropriately formalized) is true in $\mathfrak{M}$, we do *not* get from this that "$Num(\mathfrak{M})\models$ each PA axiom" is true in $\mathfrak{M}$. And this is just like how being able to check each individual derivation in PA doesn't give us a way to check all derivations at once, so it really shouldn't be surprising.