Assume the square of sum of $S$ of $v_1, v_2, \dots, v_N$ means the square of the sum of all possible combinations of $S$ values among $v_1, v_2, \dots, v_N$, i.e., $$ \sum_{i=1}^{{N \choose S}} (\sum_j v_j)^2 $$ where the inner sum has $S$ values.
My try
This sum can be written as the following: $$ \alpha(v_1^2+\dots+ v_N^2)+\beta(v_1v_2+\dots+v_{N-1}v_N). $$
Question
Is there an intuitive way of finding $\alpha$ and $\beta$?
As requested, I am expanding my comment into an answer. This is a symmetric polynomial, so it is enough to find out how many times $v_1^2$ and $v_1v_2$ occur in the sum.
The square $v_1^2$ occurs whenever a set contains $1$ and $S−1$ arbitrary other elements in $\{2,\dots, N\}$. Therefore, $$\alpha=\binom{N-1}{S-1}=\frac{S}{N}\binom{N}{S}.$$ The product $v_1v_2$ occurs twice in every sum over a set of $S$ elements containing $v_1$ and $v_2$. The other $S−2$ out of $N−2$ elements in $\{3,\dots, N\}$ are arbitrary, so $$\beta=2\binom{N-2}{S-2}=2\frac{S(S-1)}{N(N-1)}\binom{N}{S}=2\frac{\binom{S}{2}}{\binom{N}{2}}\binom{N}{S}.$$ And indeed, you can check that the number of terms in the sum is $$\alpha N+\beta\binom{N}{2}=N\binom{N-1}{S-1}+N(N−1)\binom{N-2}{S-2}=(S+S(S−1))\binom{N}{S}=S^2\binom{N}{S},$$ which is as expected, since every summand in $$\sum_{\substack{A\subseteq \{1,\dots,N\}\\|A|=S}}{\left(\sum_{j\in A}v_j\right)^2}$$ contains $\binom{N}{S}$ squares of sums of $S$ summands each.