Clarification in a particular application of Le Chatelier principle
For starters, I wouldn't classify your equilibrium as a buffer. A buffer requires excess weak acid, and it also requires either a strong base or the salt of the weak acid (which you don't have). The equilibrium you stated is the dissociation of CH3COOH into its conjugate base and a conjugate acid.



Say you had the following buffer:

$$CH_{3}COOH + NaOH ⇌ CH_{3}COONa + H_{2}O$$

The equilibrium of this would be:

$$CH_{3}COOH ⇌ CH_{3}COO^{-} + H^{+}$$

And the ionic equation would be:

$$H^{+} + OH^{-} → H_{2}O$$

As per definition, a buffer is a system that minimises pH change on small additions of acid or alkali. How? All down to equilibrium...

If you were to increase $[H^{+}]$:


In terms of the equilibrium, the added $H^{+}$ would react with the base, $CH_{3}COO^{-}$, to form 
$CH_{3}COOH$. This is Le Chatelier's principle in the fact that increasing $[H^{+}]$ shifts the equilibrium to the left.


If you were to increase $[OH^{-}]$:


In terms of the ionic equation, the added $OH^{-}$ would react with the $H^{+}$ to form $H_{2}O$. This is Le Chatelier's principle in the fact that increasing $[OH^{-}]$ shifts the equilibrium to the right.




In both of these, it's clear that the addition of $H^{+}$ or $OH^{-}$ would just be an acid-base reaction. The products formed minimise the $pH$ of the system due to their relative strengths compared with the added acids/alkalis.

The $pH$ does change though, but just not as much as it would if it wasn't in the buffer system. A buffer just keeps the equilibrium central.