$Q$ is a relation as described above, $\Sigma$ is consistent, and $\phi$ is a formula with one variable. I think the relation in above holds because if $c$ does not belong in $Q$, then by the relation $\Sigma$ cannot prove $\phi[c]$. Not being able to prove a sentence is not the same as proving the negation of the sentence, right?
Also, what if I have another formula $\psi(w)$ and I'm given $Q(c) \iff \Sigma \vdash \lnot \psi[c]$? Does $\lnot Q(c) \iff \Sigma \not\vdash \lnot\psi[c]$? I'm thinking it is so because I treat $\lnot \psi$ as $\phi$ and argue like before.
You are right.
If $Q(c) \iff \Sigma \vdash \phi[c]$, then obviously :
Then, "internalizing" the negations, in the LHS we have $¬Q(c)$ while in the RHS we have to "negate" the relation $\vdash$.
The same for the second example.
We have an equivalence between two statements : one is the LHS, i.e. the formula $¬Q(c)$, the other one is the "meta-statement" :
When we negate a statement we add the negation to the "verb" : in the example, the negation of $\vdash \phi$ ("$\phi$ is provable") is $\nvdash \phi$ ("$\phi$ is not provable") and not : $\vdash \lnot \phi$ ("$\lnot \phi$ is provable").
The denial of "I will eat the cake" is "I will not eat the cake" (and not : "I will eat the not-cake" ...)