Question 1. Let $\operatorname{con}(\mathsf{ZFC})$ be a sentence in $\mathsf{ZFC}$ asserting that $\mathsf{ZFC}$ has a model.
Let $S$ be the theory $\mathsf{ZFC}+\operatorname{con}(\mathsf{ZFC})$. Let $\operatorname{con}(S)$ be a sentence in $S$ (or $\mathsf{ZFC}$) asserting that $S$ has a model. We assume throughout that $\mathsf{ZFC}$ is consistent.
Is it provable in $\mathsf{ZFC}$: $\operatorname{con}(\mathsf{ZFC})\leftrightarrow\operatorname{con}(S)$?
Question 2. Assume now S is inconsistent i.e. that is $S$, has no model.This means that there is no model of $\mathsf{ZFC}$ in which $\operatorname{con}(\mathsf{ZFC})$ is true, so the negation of $\operatorname{con}(\mathsf{ZFC})$ is true in any model of $\mathsf{ZFC}$ and therefore (by completeness theorem) is a theorem of $\mathsf{ZFC}$?
No, not at all.
The consistency of $S$ is a strictly stronger assertion than the consistency of $\sf ZFC$. Note that $S$ proves $\operatorname{Con}\sf(ZFC)$ but not $\operatorname{Con}(S)$ (as per the usual incompleteness arguments).
If $\sf ZFC$ would prove the equivalence between the two consistency statements, then $S$ would prove that, in which case $S$ would prove its own consistency, because it can prove $\operatorname{Con}\sf(ZFC)$.
To the edit,
Note that you do have an additional assumption in the meta-theory. So now we actually have $\sf M=T+\operatorname{Con}(ZFC)+\lnot\operatorname{Con}(ZFC+\operatorname{Con}(ZFC))$ as our meta-theory (where $\sf T$ is some sufficient theory e.g. $\sf ZFC$).
This means that $M$ proves that $\sf ZFC$ proves that $\lnot\operatorname{Con}\sf(ZFC)$. It doesn't mean that $\sf ZFC$ proves that directly. The proof lies in the two additional consistency assumptions, rather than the part of $\sf ZFC$.