Proof that all Boolean functions can be computed by $(MOD_2-MOD_3)$ circuit
The key observation is that $f$ is not represented by any CNF but by a CNF where all clauses have $n$ literals with distinct variables where $n$ is the number of variables.

It is easy to see that such a CNF exists. Indeed, let $S$ be the set of solutions of $\neg f$. Each solution $s \in S$ can be seen as a conjunction of $n$ literals. For example, $s = \{x=0, y=1, z=0\}$ may be seen as $\neg x \wedge y \wedge \neg z$. Thus $\neg f = \bigvee_{s \in S} s$ and then $f = \bigwedge_{s \in S} \neg s$ which is a CNF formula.

Now this is important since the clauses of this CNF, let's call it $F$, are  in one to one correspondance with assignments of the variables: to each clause $\neg s$ of $F$, we associate $s$, the only assignment that does not satisfy the clause. For instance, the only way not to satisfy  $\neg s = x \vee \neg y \vee z$ is $s=\{x=0, y=1, z=0\}$. Thus, given an assignment, at most one clause of $F$ is not satisfied.

Now there is still something to explain. The authors claim that we can assume $m \equiv 1 \mod 3$ wlog. That is not completely true since we cannot grow the number of clauses artificially and still have the previous property. However, for the sake of their argument, it is enough to add trivial clauses such a $x \vee \neg x$ that always evaluate to $1$.