A proof of Liouville’s theorem I have found a proof of Liouville's theorem on the internet, which fits me very well except one step I don't understand, the derivation is as follows:

In the derivative, it must have used the relation $dq'_i=dq_i+\frac{\partial\dot{q}_i}{\partial q_i}dq_idt$ and $dp'_i=dp_i+\frac{\partial\dot{p}_i}{\partial p_i}dp_idt$  which I don't understand. Since in the Hamiltonian's formalism, the independent variables are $q$'s and $p$'s, why $dq_i'$ not equal to $dq_i+\sum_j\frac{\partial\dot{q}_i}{\partial q_j}dq_jdt + \sum_j\frac{\partial\dot{q}_i}{\partial p_j}dp_jdt$, etc?
 A: Consider 
\begin{equation}
d\Gamma' 
= \left( dq_1 + \frac{\partial\dot{q}_1}{\partial q_1}dq_1dt + \cdots  
        + \frac{\partial\dot{q}_1}{\partial p_1}dp_1dt 
        + \cdots
 \right)
\left( dq_2 
        + \cdots
 \right) \cdots (dp_1+\cdots)(dp_{3N}+\cdots)
\end{equation}
where the omitted dots are the remaining terms in your correct expansion 
\begin{equation}
dq'_i = dq_i + \sum_j \frac{\partial\dot{q}_i}{\partial q_j}dq_j dt + \sum_j \frac{\partial\dot{q}_i}{\partial p_j} dp_jdt. 
\end{equation}
Now we want to expand the expression of $d\Gamma'$ up to the first order in $dt$. That means we can only choose one $dt$ term in each bracket. Consider we choose $dt$ factor from the first bracket. If we choose 
\begin{equation}
\frac{\partial\dot{q}_1}{\partial q_1}dq_1dt
\end{equation}
from the first bracket, then all other brackets should contribute terms like $dq_2,\cdots$ and this leads to your (2.19). 
Suppose we instead choose a term in linear $dt$ like 
\begin{equation}
\frac{\partial\dot{q}_1}{\partial p_1}dp_1dt. 
\end{equation} 
Again, other brackets should contribute something without $dt$ and these can only be $dq_2,\cdots$ and also the term $dp_1$. However, the volume form is a top form, you can't have $dp_1$ show up twice, therefore, these terms do not contribute. 
