Given: $C_N = N + 1 + \displaystyle\sum_{1\leq k \leq N} \frac{(N - k)(k - 1)}{N \choose 3}(C_{k-1} + C_{N-k})$ for $N > 2$ with $C_0 = C_1 = C_2 = 0$ (and the implied $C_3 = 4$). We follow through the standard steps for solving recurrences with generating functions.
${N \choose 3}C_N = (N + 1){N \choose 3} + 2\displaystyle\sum_{1\leq k \leq N} (N - k)(k - 1)C_{k-1}$ and my question is: was a substitution made for $C_{N-k}$ such as $C_{4-1} = C_3$ or how did the $2$ appear?
Yes, the sum over $C_{N-k}$ can be rewritten by substituting $k'=N-k+1$, giving $$ \sum_{k=1}^N (N-k)(k-1)C_{N-k} = \sum_{k'=1}^N(N-(N-k'+1))(N-k')C_{N-(N-k'+1)}= \sum_{k'=1}^N(k'-1)(N-k')C_{k'-1},$$ which is the same as the summand over $C_{k-1}$. Hence, both summands are equivalent, allowing to simplify with the factor of 2.