Note that we can better write each term as
$$
\eqalign{
& S(k,n)=\sum\limits_{j = 0}^{\left\lfloor {{{i + n - k} \over {n + 2}}} \right\rfloor } {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
n - 1 \cr} \right)} = \cr
& = \sum\limits_{0\, \le j} {\left[ {0 \le i + n - k - j\left( {n + 2} \right)} \right]\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
n - 1 \cr} \right)} = \quad \quad (1) \cr
& = \sum\limits_{0\, \le j} {\left[ {0 \le i + n - k - j\left( {n + 2} \right)} \right]\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} = \quad \quad (2) \cr
& = \left[ {1 \le n} \right]\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} \quad \quad (3) \cr}
$$
where $[P]$ denotes the Iverson bracket
and where:
- (1) we replace the upper bound with Iverson bracket;
- (2) we can apply reflection, since the upper term of the binomial is non-negative;
- (3) we can omit the Iverson bracket, because for $1 \le n$ it is implicit in the binomial.
Note that $S(k,n)$ is equivalent to
$$
S(k,n) = N_{\,b} (i + 1 - k,\,n + 1,n)
$$
where
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{k}
\binom
{ s + m - 1 - k\left( {r + 1} \right) }
{ s - k\left( {r + 1} \right)}\ }
$$
is the number discussed in this related post, and is also called the "r-nomial coefficient" since
$$
F_b (x,r,m) = \sum\limits_{0\,\, \leqslant \,\,s\,\,\left( { \leqslant \,\,r\,m} \right)} {N_b (s,r,m)\;x^{\,s} }
= \left( {1 + x + \cdots + x^{\,r} } \right)^m = \left( {\frac{{1 - x^{\,r + 1} }}{{1 - x}}} \right)^m
$$
Now, the cumulative sum of $N_b$ has a similar expression
$$
M_{\,b} (t,\,r,m) = \sum\limits_{0\, \le s\; \le \,\,t} {N_{\,b} (s,\,r,m)}
= \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{t \over {r + 1}}\, \le \,m} \right)} {\left( { - 1} \right)^k \binom {m}{k}
\binom{ t + m - k\left( {r + 1} \right)} {t - k\left( {r + 1} \right) } }
$$
but this does not help to simplify much your sum, since it will always leave three terms.
The attempt to directly perform the sum will give:
$$
\eqalign{
& S(n) = \left[ {1 \le n} \right]\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( \matrix{
2 \cr
k \cr} \right)\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
2 \cr
k \cr} \right)\left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,k + j} \left( \matrix{
k - 3 \cr
k \cr} \right)\left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - \left( {k + j} \right) - j\left( {n + 1} \right) \cr
i + 1 - \left( {k + j} \right) - j\left( {n + 1} \right) \cr} \right)} } = \cr
& = \sum\limits_{0\, \le l} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,l} \left( \matrix{
l - j - 3 \cr
l - j \cr} \right)\left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - l - j\left( {n + 1} \right) \cr
i + 1 - l - j\left( {n + 1} \right) \cr} \right)} } \cr}
$$
and we cannot proceed to simplify by applying double correlation, due to the presence of $\left( { - 1} \right)^{\,l}$
We could instead simplify the related sum
$$
\eqalign{
& S_{\, - } (n) = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
2 \cr
k \cr} \right)S(k,n)} = \cr
& = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
2 \cr
k \cr} \right)\sum\limits_{} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( \matrix{
k - 3 \cr
k \cr} \right)\left( \matrix{
i + n - k - j\left( {n + 2} \right) \cr
i + 1 - k - j\left( {n + 2} \right) \cr} \right)} } = \cr
& = \sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
n \cr
j \cr} \right)\left( \matrix{
i + n - 2 - j\left( {n + 2} \right) \cr
i + 1 - j\left( {n + 2} \right) \cr} \right)} \cr}
$$
so that the actual sum may be reduced to two terms.
Besides these, and other related manipulations, I do not see better ways.