Why can we use a weaker metatheory for consistency results? I'm not entirely sure why we can get away with using weaker metatheories for consistency results.
For instance consider the following, $ZFC \not \vdash Con(ZFC) \rightarrow Con(ZFC+I)$. This is because $ZFC + I \vdash Con(ZFC)$, and then if $ZFC \vdash Con(ZFC) \rightarrow Con(ZFC + I)$, we'd get $ZFC + I \vdash Con(ZFC + I)$, contradicting Godel's incompleteness theorem. So $ZFC$ can't be a metatheory for proving $Con(ZFC) \implies Con(ZFC + I)$. But $ZFC$ is far stronger than $PA$, so if it can't prove the consistency statement, why can $PA$? For instance, if $PA \vdash Con(ZFC) \implies Con(ZFC + I)$, then obviously $ZFC + I \vdash Con(PA)$, so why doesn't the same contradiction as above occur? Is it that $ZFC + I$ doesn't necessarily prove $Con(ZFC) \rightarrow Con(ZFC + I)$ even if $PA \vdash Con(ZFC) \rightarrow Con(ZFC+I)$? Then why did it work in the contradiction above? Was it because $ZFC \subseteq ZFC + I$?
This is just a specific example, but my question in general is: it seems that we can't prove consistency results in fairly strong theories, yet I often hear people say that $PA$ or even a finite fragment of $PA$ is enough for most consistency proofs.
Another example: using $ZFC + Con(ZFC) + \neg Con(ZFC + $"Inaccessibles exist"$)$ as our metatheory we definitely can't prove $Con(ZFC) \rightarrow Con(ZFC + $"inaccessibles exist"$)$. But the latter is a fairly easy consistency proof which we say that can be done in fairly weak metatheories. $ZFC + Con(ZFC) + \neg Con(ZFC + $"inacessibles exist"$)$ seems pretty strong to me.
Overall I'm interested in how the metatheory comes into play when describing consistency statements; it seems to me that the metatheory is very much important and yet people usually don't even mention what they're using.
 A: I think this question is based on an error, namely the claim that $Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$ is considered a theorem of the "standard metatheory" of mathematics. This is not the case. In particular, when people (correctly!) say that most relative consistency theorems can be proved in $\mathsf{PA}$ or even weaker systems, there's no tension here since $Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$ is not a relative consistency theorem in the sense meant.
Specifically, here's how I would phrase the heuristic claim above:

In practice so far, whenever $T,S$ are "naturally occurring" theories containing $\mathsf{ZF}$ with $\mathsf{ZF}\vdash Con(T)\rightarrow Con(S)$, we have $\mathsf{PA}\vdash Con(T)\rightarrow Con(S)$.

(And in fact even less than $\mathsf{PA}$ is needed here; I'm not an expert here, but my understanding is that we can go down to $\mathsf{I\Sigma_1}$ without serious effort and even further if we take more care. Hajek/Pudlak's book is a good resource here if you're interested.)
Note that the clause "containing $\mathsf{ZF}$" there is crucial, since if we go below $\mathsf{ZF}$ things trivialize: e.g. we boringly have $$\mathsf{ZF}\vdash Con(\emptyset)\rightarrow Con(\mathsf{PA})$$ since $\mathsf{ZF}\vdash Con(\mathsf{PA})$ outright, but of course $$\mathsf{PA}\not\vdash Con(\emptyset)\rightarrow Con(\mathsf{PA}).$$
The point then is that $T=\mathsf{ZFC}, S=\mathsf{ZFC+I}$ does not constitute a counterexample to this heuristic since - as you observe - we don't have $\mathsf{ZF}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC+I})$ (unless $\mathsf{ZF}$ is inconsistent that is).

As to justifying stronger systems from below, Godel's second incompleteness theorem - as you observe in the OP - shows that we cannot reach $\mathsf{ZFC+I}$ from $\mathsf{ZFC}$ in the same way that we can reach $\mathsf{ZFC}$ from $\mathsf{ZF}$ (or $\mathsf{ZFC+GCH}$ from $\mathsf{ZF}$, or similar). The relevant term here is consistency strength: $\mathsf{ZFC+I}$ has strictly greater consistency strength than $\mathsf{ZFC}$ (unless the latter is inconsistent of course).
