Notice the "modular" factorization (keep in mind $n=m_1+m_2+\cdots+m_r$):
$[n]_q=[m_1]_q+q^{m_1}[m_2]_q+q^{m_1+m_2}[m_3]_q+\cdots+q^{m_1+\cdots+m_{r-1}}[m_r]_q. \tag{a}$
Basically what we've done is taken $[n]_q=q^0+q^1+q^2+\cdots+q^{n-1}$ and partitioned it into the first $m_1$ terms, the second $m_2$ terms, and so on to the last $m_r$ terms, and then factored out the highest power of $q$ from each cell of the partition until we're left with a linear combination of the analog terms $[m_i]_q$, $i=1,\dots,r$. Now take $(a)$ and divide both sides by $[n]_q$, then multiply by the $q$-multinomial and distribute the multiplication through to obtain
$\frac{[m_1]_q}{[n]_q}{n\choose m_1,\dots,m_r}+\cdots+\frac{q^{m_1+\cdots+m_{r-1}}[m_r]_q}{[n]_q}{n\choose m_1,\dots,m_r}. \tag{b}$
Use the relation $[a]_q!=[a]_q\cdot[a-1]_q$ in the $q$-factorials implicit in each term's $q$-multinomial;
$\small \binom{n}{m_1,m_2,\dots,m_r}_q=\binom{n-1}{m_1-1,m_2,\dots,m_r}_q+\cdots q^{m_1+\cdots+m_{r-1}}\binom{n-1}{m_1,m_2,\dots,m_r-1}_q.$
Here's an example to better illustrate $(a)$. Set $n=9, m_{1,2,3}=2,3,4$. Then
$[9]_q=\color{Red}{1+q}+\color{Green}{q^2+q^3+q^4}+\color{Blue}{q^5+q^6+q^7+q^8}$
$=(\color{Red}{1+q})+q^2(\color{Green}{1+q+q^2})+q^5(\color{Blue}{1+q+q^2+q^3})$
$=[2]_q+q^2[3]_q+q^{2+3}[4]_q.$