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First off, there should be $(-1)^{i+1}$ not $(-1)^{R+i}$ (now it's corrected in the question).
UPDATE. Argument below is simplified and streamlined.

The identity generalizes the previous question, which essentially represents the case of $R=1$. In fact, this generalized identity has a somewhat easier proof (given below) than my earlier proof for $R=1$.

Again, we start with noticing that the inner sum equals $i!\cdot S(2k+1,i+1)$, where $S(\cdot,\cdot)$ are Stirling numbers of second kind.

Let $k>0$ is fixed, and $LHS(2k,n,R)$ denote the identity left-hand side, i.e. $$LHS(2k,n,R) := \sum_{i=0}^{2k} \big( \binom{n+R-1}{R+i} + (-1)^{i+1} \binom{n+R+i}{R+i}\big) i! S(2k+1,i+1).$$

We start to prove the identity in two cases: $R=0$ and $n=0$. In the case $R=0$, we have \begin{split} LHS(2k,n,0) &= \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} In the case $n=0$ and $R>0$, we have \begin{split} LHS(2k,0,R) &= \sum_{i=0}^{2k} (-1)^{i+1} i! S(2k+1,i+1) \\ &= -\sum_{i=0}^{2k} (-1)_i S(2k+1,i+1) \\ &= -0^{2k} = 0.\end{split}

Now, we are ready to prove the identity by induction on $n+R$. From above, it follows that $L(2k,n,R)=0$ when $(n,R)=(0,1)$ or $(n,R)=(1,0)$, i.e. when $n+R=1$. Now, when $n+R>1$, the cases $n=0$ or $R=0$ are addressed above, while in the case $n>0$ and $R>0$, we use Pascal's rule to conclude that $$LHS(2k,n,R) = LHS(2k,n,R-1) + LHS(2k,n-1,R) = 0$$ by the induction assumption.

First off, there should be $(-1)^{i+1}$ not $(-1)^{R+i}$ (now it's corrected in the question).
UPDATE. Argument below is simplified and streamlined.

The identity generalizes the previous question, which essentially represents the case of $R=1$. In fact, this generalized identity has a somewhat easier proof (given below) than my earlier proof for $R=1$.

Again, we start with noticing that the inner sum equals $i!\cdot S(2k+1,i+1)$, where $S(\cdot,\cdot)$ are Stirling numbers of second kind.

Let $k>0$ is fixed, and $LHS(2k,n,R)$ denote the identity left-hand side, i.e. $$LHS(2k,n,R) := \sum_{i=0}^{2k} \big( \binom{n+R-1}{R+i} + (-1)^{i+1} \binom{n+R+i}{R+i}\big) i! S(2k+1,i+1).$$

We start to prove the identity in two cases: $R=0$ and $n=0$. In the case $R=0$, we have \begin{split} LHS(2k,n,0) &= \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} In the case $n=0$ and $R>0$, we have \begin{split} LHS(2k,0,R) &= \sum_{i=0}^{2k} (-1)^{i+1} i! S(2k+1,i+1) \\ &= -\sum_{i=0}^{2k} (-1)_i S(2k+1,i+1) \\ &= -0^{2k} = 0.\end{split}

Now, we are ready to prove the identity by induction on $n+R$. From above, it follows that $LHS(2k,n,R)=0$ when $(n,R)=(0,1)$ or $(n,R)=(1,0)$, i.e. when $n+R=1$. Now, when $n+R>1$, the cases $n=0$ or $R=0$ are addressed above, while in the case $n>0$ and $R>0$, we use Pascal's rule to conclude that $$LHS(2k,n,R) = LHS(2k,n,R-1) + LHS(2k,n-1,R) = 0$$ by the induction assumption.

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.

First off, there should be $(-1)^{i+1}$ not $(-1)^{R+i}$ (now it's corrected in the question).
UPDATE. Argument below is simplified and streamlined.

The identity generalizes the previous question, which essentially represents the case of $R=1$. In fact, this generalized identity has a somewhat easier proof (given below) than my earlier proof for $R=1$.

Again, we start with noticing that the inner sum equals $i!\cdot S(2k+1,i+1)$, where $S(\cdot,\cdot)$ are Stirling numbers of second kind.

Let $k>0$ is fixed, and $LHS(2k,n,R)$ denote the identity left-hand side, i.e. $$LHS(2k,n,R) := \sum_{i=0}^{2k} \big( \binom{n+R-1}{R+i} + (-1)^{i+1} \binom{n+R+i}{R+i}\big) i! S(2k+1,i+1).$$

We start to prove the identity in two cases: $R=0$ and $n=0$. In the case $R=0$, we have \begin{split} LHS(2k,n,0) &= \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} In the case $n=0$ and $R>0$, we have \begin{split} LHS(2k,0,R) &= \sum_{i=0}^{2k} (-1)^{i+1} i! S(2k+1,i+1) \\ &= -\sum_{i=0}^{2k} (-1)_i S(2k+1,i+1) \\ &= -0^{2k} = 0.\end{split}

Now, we are ready to prove the identity by induction on $n+R$. From above, it follows that $L(2k,n,R)=0$ when $(n,R)=(0,1)$ or $(n,R)=(1,0)$, i.e. when $n+R=1$. Now, when $n+R>1$, the cases $n=0$ or $R=0$ are addressed above, while in the case $n>0$ and $R>0$, we use Pascal's rule to conclude that $$LHS(2k,n,R) = LHS(2k,n,R-1) + LHS(2k,n-1,R) = 0$$ by the induction assumption.

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, having the identity established for $R=0$ and $R=1$, for larger $R$ it follows by induction thanks to Pascal's rule.

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.