Here we look at the connection between binomial expansion and multinomial expansion for the cases $n=2$ and $n=3$ which might give a better idea what's going on.
Case $n=2$:
\begin{align*}
\color{blue}{(x_1+x_2)^n}&\color{blue}{=\sum_{k_1=0}^n\binom{n}{k_1}x_1^{k_1}x_2^{n-k_1}}\\
&=\sum_{k_1=0}^n\frac{n!}{k_1!(n-k_1)!}x_1^{x_1}x_2^{n-k_1}\\
&=\sum_{{k_1+k_2=n}\atop{k_1,k_2\geq 0}}\frac{n!}{k_1!k_2!}x_1^{k_1}x_2^{k_2}\tag{1}\\
&\,\,\color{blue}{=\sum_{{k_1+k_2=n}\atop{k_1,k_2\geq 0}}\binom{n}{k_1,k_2}x_1^{k_1}x_2^{k_2}}
\end{align*}
Comment:
- In (1) we introduce a new index variable $k_2=n-k_1$. Note we also state in the index region $k_1,k_2\geq 0$ which is sometimes silently assumed.
Case $n=3$:
\begin{align*}
\color{blue}{(x_1+x_2+x_3)^n}&=(x_1+(x_2+x_3))^n\\
&=\sum_{k_1=0}^n\binom{n}{k_1}x_1^{k_1}(x_2+x_3)^{n-k_1}\\
&\,\,\color{blue}{=\sum_{k_1=0}^n\binom{n}{k_1}x_1^{k_1}\sum_{k_2=0}^{n-k_1}\binom{n-k_1}{k_2}x_2^{k_2}x_3^{n-k_1-k_2}}\\
&=\sum_{k_1=0}^n\sum_{k_2=0}^{n-k_1}\frac{n!}{k_1!(n-k_1)!}\cdot\frac{(n-k_1)!}{k_2!(n-k_1-k_2)!}x_1^{k_1}x_2^{k_2}x_3^{n-k_1-k_2}\\
&=\sum_{k_1=0}^n\sum_{k_2=0}^{n-k_1}\frac{n!}{k_1!k_2!(n-k_1-k_2)!}x_1^{k_1}x_2^{k_2}x_3^{n-k_1-k_2}\\
&=\sum_{k_1=0}^n\sum_{{k_2+k_3=n-k_1}\atop{k_2,k_3\geq 0}}\frac{n!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}\tag{2}\\
&=\sum_{{k_1+k_2+k_3=n}\atop{k_1,k_2,k_3\geq 0}}\frac{n!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}\\
&\,\,\color{blue}{=\sum_{{k_1+k_2+k_3=n}\atop{k_1,k_2,k_3\geq 0}}\binom{n}{k_1,k_2,k_3}x_1^{k_1}x_2^{k_2}x_3^{k_3}}\\
\end{align*}
- In (2) we introduce a new index variable $k_3=n-k_1-k_2$ similarly as we did in (1).
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.