In the comments of Yuval Filmus' answer, the OP suggested that while, for any two CNF formulas $f$ and $\tilde{f}$,
($f \text{ falsifiable} \iff \tilde{f} \text{ satisfiable}) \iff (f \text{ satisfiable} \iff \tilde{f} \text{ falsifiable}$)
wasn't usually true, it was implied in that specific case by the parsimonious reduction of the CNFFAL instance $f$ to the CNFSAT instance $\tilde{f}$, thus proving that
$f \text{ satisfiable} \iff \Phi_{NTM_{CNFFAL},f} \text{ falsifiable}$.
This answer is supposed to show evidence that it isn't the case (assuming it isn't clear to the reader), as a complement of the previously mentioned answer (per request of the OP in the comments).
I'll try to exhibit an example making clear that it isn't the case. Now, building the corresponding $\Phi_{NTM_{CNFFAL},f}$ of some $f$ would be a bit tedious, so I'll build a parsimonious reduction from CNFFAL to SAT instead (contrast with the parsimonious reduction from CNFFAL to CNFSAT used in the question), but such that the falsifiability of the SAT instance doesn't imply the satisfiability of the CNFFAL instance.
I claim that, for any CNF formula $f$,
any assignment $(a_1, \dots, a_n)$ falsifying $f(a_1, \dots, a_n)$ allows me to build a unique satisfying assignment of $\tilde{f}(a_1, \dots, a_n, c) = \lnot f(a_1, \dots, a_n) \land c$ (and vice-versa). This (admittedly dumb) reduction from CNFFAL to SAT preserves the number of solutions so it's parsimonious.
Now let's assume $f(a, b) = (a \lor b) \land (\lnot a \lor b) \land (a \lor \lnot b) \land (\lnot a \lor \lnot b)$.
There indeed are as many ways to falsify $f$ as to satisfy $\tilde{f}$ (4, per the previous parsimonious reduction). Therefore:
$f \text{ falsifiable} \iff \tilde{f} \text{ satisfiable}$
However there also are 4 ways to falsify $\tilde{f}$ (just set $c$ to 0) whereas $f$ is clearly unsatisfiable. Therefore:
$\lnot(f \text{ satisfiable} \iff \tilde{f} \text{ falsfiable})$
Contrast this with that excerpt of the reasoning in the question:
Therefore $\Phi_{NTM_{CNFFAL},f}\equiv\lnot f$ since Cook and Levin reduction is parsimonious and $\Phi_{NTM_{CNFFAL},f}$ is also in conjunctive normal form.
Therefore f is falsifiable if and only if $\Phi_{NTM_{CNFFAL},f}$ is satisfiable.
Therefore f is satisfiable if and only if $\Phi_{NTM_{CNFFAL},f}$ is falsifiable.
The previous statement stems from the fact that $\tilde{f}$ uses more variables than $f$, hence the number of possible assignments of $\tilde{f}$ is greater than the number of possible assignments of $f$. Indeed, if those two numbers of possibilities were equal, the reduction being parsimonious would indeed lead to the equivalence of $f$ being satisfiable and $\tilde{f}$ being falsifiable (by a simple counting argument, as the set of falsifiable assignments is the complement of the set of satisfiable assignments).
In the question, $\Phi_{NTM_{CNFFAL},f}$ has been obtained thanks to Cook-Levin's reduction from an arbitrary NP problem to CNFSAT. That formula doesn't necessarily requires the same amount of variable as $f$ (and is actually very likely to require a far greater amount), so the equivalence doesn't necessarily holds.