From the two-point forward difference formula, we have:
$$\tag 1 f'(x) \approx \dfrac{f(x+h)-f(x)}{h}$$
Using Richardson's, we replace $h$ by $\dfrac{h}{2}$ in $(1)$, yielding:
$$\tag 2 f'(x) \approx 2\dfrac{f(x+h/2)-f(x)}{h}$$
Subtracting $2 \times (2)$ from $(1)$ (this is to get the Richardson form) yields:
$$f'(x) \approx \frac{- f(x+h) + 4 f(x+h/2)- 3f(x) }{h} $$
This is a $3-$point, $2^{nd}$ order approximation.