Let $h(x)=f(x)g(x)$, and let the $\Delta$ be the forward difference operator on functions be $$\Delta f := \frac{f_{j+1}-f_j}{\Delta x}$$ where $f_j = f(j\Delta x)$
Now I can derive the product rule for finite differences in terms of $\Delta f$ and $\Delta g$ easily
$$\Delta h = \frac{h_{j+1}-h_j}{\Delta x} = \frac{f_{j+1}g_{j+1}-f_jg_j}{\Delta x} = \frac{f_{j+1}g_{j+1}\color{red}{-f_jg_{j+1} }{\color{blue}{+f_jg_{j+1}}}-f_jg_j}{\Delta x}=g_{j+1}\Delta f+f_j\Delta g$$
However, you can see in the first term that $g$ is shifted forward by $1$ step.
But on Wikipedia for the finite differences the rule is express with a second order term:
$$\Delta(fg) = f\Delta g + g \Delta f + \Delta f \Delta g$$
I have tried converting my expression to this with all sorts of algebra, and I even tried using the wiki's formula to go backwards to my equation, but nothing has worked. It's clear there must be some clever trick to convert the shift operator in terms of $f,\Delta f,\Delta g$ but I can't see how.
Any ideas?
You have $g_{j+1}\Delta f=(g_{j+1}-g_j)\Delta f+g_j\Delta f$, so this is the same as $g_j\Delta f+\Delta x\Delta g\Delta f$.