Ok, so I am going with Wikipedia's definition of the Richardson Extrapolation, rather than the recursive definition that I know is floating out there. One step at a time.
Given $A(h)$ is an estimate of $A^*$ with order $n$ convergence, an improved estimate can be found using the following:
$$R(h,p) = \frac{p^nA(\frac{h}{p}) - A(h)}{p^n - 1}$$
I believe I understood the derivation, but the application is elluding me, so I decided to try finding Simpson's Rule using the composite Trapezoid Rule:
$$A(h/2) = \frac{h}{2}\bigg(\frac{f(x_0) + f(x_N)}{2} + \sum^{N-1}_{k=1}f(x_k)\bigg) $$
$$A(h) = h\bigg(\frac{f(x_0) + f(x_N)}{2} + \sum^{N/2-1}_{k=1}f(x_{2k})\bigg) $$
Then
$$ \begin{align} R(h,2) &= \bigg[2h\bigg(\frac{f(x_0) + f(x_N)}{2} + \sum^{N-1}_{k=1}f(x_k)\bigg) - h\bigg(\frac{f(x_0) + f(x_N)}{2} + \sum^{N-1}_{k=1}f(x_k)\bigg) \bigg] \frac{1}{3} \\\ &= \frac{h}{3}\bigg( f(x_0) + f(x_N) + 2\sum_{k = 1}^{N -1}f(x_k) - \frac{f(x_0)}{2} - \frac{f(x_N)}{2} - \sum f(x_k) \bigg) \\\ &= \frac{h}{3}\bigg(\frac{f(x_0)}{2} + 2f(x_1) + f(x_2) + \cdots + f(x_{N-2}) + 2f(x_{N-1}) + \frac{f(x_N)}{2}\bigg) \\\ & = \frac{h}{6}\bigg(f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 2f(x_{N-2}) + 4f(x_{N-1}) + f(x_N)\bigg) \end{align} $$
EDIT: Thanks to at Lutz' suggestion, I found half the culprit, and have edited my question. However, I am left with a factor of $h/6$ out front, that I believe is still incorrect.