Ok since whuber commentedcommented, I decided to look back into it and figured it out.
I think this definitionthe following expression from wikipedia makes it easier to understand: $$ \sum_{k=0}^n f_k g_k = f_0 \sum_{k=0}^n g_k + \sum_{j=0}^{n-1} (f_{j+1} - f_j) \sum_{k=j+1}^n g_k $$$$ \sum_{k=0}^n f_k g_k = f_0 \sum_{k=0}^n g_k + \sum_{j=0}^{n-1} (f_{j+1} - f_j) \sum_{k=j+1}^n g_k. $$
Also as whuber hinted:
$$ f_k = \sum_{i'=0}^{k-1} 1 $$$$ f_k = \sum_{i'=0}^{k-1} 1 . $$
So the first part of the definition is $$ \sum_{k=0}^n f_k g_k = \sum_{k=0}^{\infty} ( P(X=k) * \sum_{i'=0}^{k-1} 1 ) $$$$ \sum_{k=0}^n f_k g_k = \sum_{k=0}^{\infty}\left ( P(X=k) \times \sum_{i'=0}^{k-1} 1 \right) $$
So the equivalent is,
$$ f_0 \sum_{k=0}^n g_k + \sum_{j=0}^{n-1} (f_{j+1} - f_j) \sum_{k=j+1}^n g_k =\sum_{i'=0}^{0} 1 * \sum_{k=0}^{\infty} P(X=k) + \sum_{j=0}^{\infty} (\sum_{i'=0}^{j+1} 1 - \sum_{i''=0}^{j} 1 ) \sum_{k=j+1}^n P(X=k) $$$$ f_0 \sum_{k=0}^n g_k + \sum_{j=0}^{n-1} (f_{j+1} - f_j) \sum_{k=j+1}^n g_k =\sum_{i'=0}^{0} 1 \times \sum_{k=0}^{\infty} P(X=k) + \sum_{j=0}^{\infty} \left(\sum_{i'=0}^{j+1} 1 - \sum_{i''=0}^{j} 1 \right) \sum_{k=j+1}^n P(X=k) $$
$$ = 0 + \sum_{j=0}^{\infty} 1 * \sum_{k=j+1}^{\infty} P(X>j) = \sum_{j=0}^{\infty} P(X>j) = \sum_{j=0}^{\infty} 1 - P(X<j) $$$$ = 0 + \sum_{j=0}^{\infty} 1 \times \sum_{k=j+1}^{\infty} P(X>j) = \sum_{j=0}^{\infty} P(X>j) = \sum_{j=0}^{\infty} 1 - P(X<j). $$
