Why doesn't a buffer solution change ph(Appreciably?)
Having problems with that concept myself, above answers seem mighty complicated, this is how I think it works (warning- only my understanding, could well be wrong)

Point One:
You have a lot of $\ce{CH3COOH}$ (not much decomposition) and a lot of $\ce{CH3COO−}$ (from the salt) and very little $\ce{H+}$

Point Two:
pH works on the amount of $\ce{OH-}$ and $\ce{H+}$ in the system, there are usually only very very small amounts (when looking at weak acids and bases) present compared to the reactant (in water, there is about 500 million molecules to ever one that dissociates) So the point I'm making is adding a few $\ce{OH-}$ or $\ce{H+}$ ions will have a big impact on the pH.

Conclusion drawn:
As you have so much $\ce{CH3COO−}$ to begin with, removing a very very small amount by adding a few $\ce{H+}$ ions and forming a very very small amount of $\ce{CH3COOH}$, is not going to have a noticeable impact on the concentrations, in comparison to the amount already in the solution (both $\ce{CH3COO−}$ and $\ce{CH3COOH}$). You could even say that it is a negligible amount and can be ignored, therefore can say this will not have an effect on the equilibrium. 

As the $\ce{H+}$ is removed, the pH is not changed; however as there is so little $\ce{H+}$ ions to begin with, not removing them will have an impact on the pH (see point two).

As already stated, there are so little $\ce{H+}$ ions in the system, removing them (by adding $\ce{OH-}$) will have an impact on equilibrium, you cannot say the amount of $\ce{H+}$ removed is negligible, you cannot ignore Le Chatelier's principle, the equilibrium will shift right to replace the lost $\ce{H+}$.

Note: I'm an A-level student, don't take this as 100% right or even 1% right, this, at best is a calculated guess as to why equilibrium can be ignored when you add $\ce{H+}$ ions and equilibrium can not be ignored when you add $\ce{OH-}$.