First off, there should be $(-1)^{i+1}$ not $(-1)^{R+i}$ (now it's corrected in the question).
The identity can be derived from my answer to the previous question, which essentially represents the case of $R=1$.
We will also need the case $R=0$, for which we have \begin{split} &\,\sum_{i=0}^{2k} \big( \binom{n-1}i + (-1)^{i+1} \binom{n+i}i\big) i! S(2k+1,i+1) \\ =&\, \sum_{i=0}^{2k} \big( (n-1)_i - (-n-1)_i \big) S(2k+1,i+1) \\ = &\, n^{2k} - (-n)^{2k} = 0. \end{split}
Now, for a fixed $k$, the identity can be proved by induction on $n+R$ as follows. Let $LHS(2k,n,R)$ denote the identity left-hand side. We proved that $L(2k,n,R)=0$ for $R=0$ and $R=1$, and thus for $n+R=1$. The Pascal's rule implies that $$LHS(2k,n,R) = LHS(2k,n,R-1) + LHS(2k,n-1,R),$$ which enables us to perform the inductive step.