Why not include another assumption within this natural deduction proof? Question: Prove $p → q \vdash ¬q → ¬p$ is valid.
The answer is:

$1. p → q~~~~\textsf{premise}$
$2. ¬q~~~~~\textsf{assumption}$
$3. ¬p~~~~~\textsf{MT }1,2$
$4. ¬q → ¬p~~~~~→\textsf{intro} 2,3$

Why isn't the assumption of $q$ included since $q$ or $¬q$ are both assumptions?

The reason we assume $\neg q$ is that we can then use it , the premise, and the modus ponens rule of inference to obtain a consequence, and on discharging the assumption we demonstrate that the premise entails the required conclusion.
$$\begin{align} & \begin{array}{l|l:l}
1  & p\to q & \textsf{premise}
\\ \hdashline 2 & \quad \neg q & \textsf{assume}
\\ 3 & \quad \neg p & \textsf{modus tollens } 1, 2
\\ \hline 4 & \neg q \to \neg p & \to\textsf{intoduction }2, 3  
\end{array} \\[0ex]\hline & p\to q ~\vdash ~\neg q\to \neg p\\ \Box
\end{align}$$
The reason we do not assume $q$ is that it is useless to.   There is no rule of inference which we can apply to it and the premise to give anything of interest.
$$\begin{align} & \begin{array}{l|l:l}
1  & p\to q & \textsf{premise}
\\ \hdashline 2 & \quad q & \textsf{assume}
\\ 3 & \quad  & (p\to q), q\textsf{ does not entail anything else} 
\\ \hline 4 & q \to q & \to\textsf{intoduction }2, 3  
\end{array} \\[0ex]\hline & p\to q ~\vdash q\to q\\ \require{cancel} \xcancel\Box
\end{align}$$
