When finding the local error for a multi-step method how do I know how far to expand my Taylor series for $y(x)$ and $y'(x)$? Another way of putting this, how do I know what order to expand to $O[h^p]$?
An example being $y_{k+1}=y_{k-1}+\frac{h}{2}(f_{k+1}+2f_{k}+f_{k-1})$
You expand it until you find the lowest order (largest when $h$ is small) non-zero error term. If that order is $k$, then $k$ is the order of the approximation. In your example, the central difference is a second-order approximation (meaning the errors are $\mathcal{O}(h^3)$. Let us look at the case of a one-sided derivative.
$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+\frac{h^3}{3!}f'''(x)+\mbox{h.o.t.}$$
Then you can see that the one-sided difference
$$f'(x)-\frac{(f_{k+1}-f_{k})}{h} = f'(x)-\frac{(f(x+h)-f(x))}{h}=-\frac{h}{2}f''(x)-\frac{h^2}{3!}f'''(x)+\mbox{h.o.t.}$$ is a first order discretization.
Once you have the lowest order, there is no pointing finding higher orders. For example if you find
$$f''(x)=+\frac{h}{2}(f_{k+1}+2f_{k}+f_{k-1}) + ah^3+bh^4+\mbox{h.o.t}$$ then it really doesn't matter what $b$ is, or what the higher order terms are because as you make $h$ small, $ah^3$ will crush them.