Why is this part of the composed proposition false? Currently I'm studying logic, but I do not understand a certain step. It is the step from row $2$ to $3$. I see that $-q$ is the common factor on both sides of the middle or operator, but I did not succeed in factoring out.
\begin{equation}
\begin{aligned}
(p \wedge q) \rightarrow \neg(r \vee q) & \equiv \neg(p \wedge q) \vee \neg(r \vee q) \\
& \equiv \neg p \vee \neg q \vee(\neg r \wedge \neg q) \\
& \equiv \neg p \vee \neg q
\end{aligned}
\end{equation}
I understand that
\begin{equation}
\begin{aligned}
&\equiv \mathbb{F} \vee(p \wedge q) \\
&\equiv(p \wedge q)
\end{aligned}
\end{equation}
so that implies that
\begin{equation}
(\neg \boldsymbol{r} \wedge \neg q)
\end{equation}
is false.
Do I miss a rule or should it be obvious from the K-map?
Whole question + solution:

 A: 
$\neg p\lor \neg q\equiv \neg p \vee \neg q \vee(\neg r \wedge \neg q) \\$

Is indeed true, one direction of the equivalence is trivial, for the other direction suppose that  $\neg p \lor \neg q \lor (\neg r \land  \neg q) $ is true then we must have that at least one of the disjuncts is true. Now suppose  that $(\neg r \land \neg q$) is true then we immediately have that $\neg q$ is true and so we get $\neg p\lor \neg q$ is true. If the other disjunct is true then we have a triviality.
A: First, I want to point out a mistake in your reasoning. You say:

I understand that


\begin{equation}
\begin{aligned}
&\equiv \mathbb{F} \vee(p \wedge q) \\
&\equiv(p \wedge q)
\end{aligned}
\end{equation}


so that implies that


\begin{equation}
(\neg \boldsymbol{r} \wedge \neg q)
\end{equation}


is false.

No, that does not follow. (and obviously $\neg r \land \neg q$ is not equivalent to $False$)
Consider: $p \lor p \equiv p$ ... so by your logic, we would have to have that $p \equiv False$?
Clearly something is wrong with your logic. At its core, what you are doing is this:
We know that if $\psi \equiv False$, then $\phi \lor \psi \equiv \phi$. But the converse is not true: if $\phi \lor \psi \equiv \phi$, then $\psi \equiv False$. In effect, you are making the logical fallacy of affirming the consequent.
OK, but why does the converse not hold? What follows below will provide some further insight.
As a general helpful principle, note that $p \lor (p \land q) \equiv p$:
$$p \lor (p \land q) \equiv p$$
$$\overset{Identity}{\equiv}$$
$$(p \land \top) \lor (p \land q)$$
$$\overset{Distribution}{\equiv}$$
$$p \land (\top \lor q)$$
$$\overset{Annihilation}{\equiv}$$
$$p  \land \top$$
$$\overset{Identity}{\equiv}$$
$$p$$
That first step is a little tricky, but you can also think of this as:
$p + pq = p(1+q) = p1 = p$
This principle is so common, that is has a name:
Absorption
$p + pq = p$
and its dual: $p(p+q) = p$
So, if you apply Absorption to your statement, you get:
$$\neg p \lor \neg q \lor (\neg r \land \neg q)$$
$$\overset{Absorption}{\equiv}$$
$$\neg p \vee \neg q$$
Yeah, it's that easy. So make sure to put Absorption in your Boolean Algebra toolkit!
Also if you look at what happens in a K-Map, you'll immediately notice why it's called Absorption:  The one term is 'absorbed' by the other term: that's why it can be removed!
So, going back to your earlier mistake:  $p \lor p \equiv p$ not because $p \equiv False$, but because the second $p$ is already covered by the first $p$.  Put differently: it's not that the second $p$ is doing nothing but because the second $p$ term isn't doing anything over and above the first term.
Likewise, $\neg q \lor (\neg r \land \neg q) \equiv \neg q$ not because $\neg r \land \neg q$ is doing nothing, but because it is doing nothing over and above $\neg q$ by itself: the $\neg q$ 'covers', and thus 'absorbs' the $\neg r \land \neg q$ term
