Combinatorial Proof of Binomial-type Identity

You’re on the right track. The $k-1$ circled stars in the even-numbered positions are the boundaries between your $k$ buckets, and the other $k$ circled stars pick one specific element from each bucket. Thus, $\binom{n+k-1}{2k-1}$ counts the number of ways to divide $n$ things into $k$ buckets and choose one thing from each bucket.

Now consider a particular vector of contents, $\langle a_1,\dots,a_k\rangle$, meaning $a_i$ things in the $i$-th bucket. How many times will this vector be counted in that binomial coefficient? Once for each way of choosing one object from each bucket, and there are $a_1a_2\dots a_k$ ways to do that. Thus, the content vector $\langle a_1,\dots,a_k\rangle$ gets counted $a_1a_2\dots a_k$ times in the binomial coefficient. And on the lefthand side you’re simply summing those products over all possible content vectors.

(Interesting identity; I’d not seen it before.)

Note that $$ 1x^1+2x^2+3x^3+4x^4+\dots=\frac{x}{(1-x)^2}\tag{1} $$ If we raise $(1)$ to the $k^{\text{th}}$ power, the coefficient of $x^n$ would be the sum of all products of $k$ positive integers that sum to $n$. That is, we want to find the coefficient of $x^{n-k}$ in $(1-x)^{-2k}$: $$ \begin{align} (-1)^{n-k}\binom{-2k}{n-k} &=\binom{n+k-1}{n-k}\\[6pt] &=\binom{n+k-1}{2k-1}\tag{2} \end{align} $$