What's happening at the beginning of a weak acid titration? At the beginning of a Weak Acid/Strong Base titration curve there is a sharp decrease in $[\ce{H+}]$ and a sharp increase in pH before the ratio between the weak acid and it's conjugate base becomes low enough that we have a buffer solution.



What's happening at that point?

My interpretation is that, given the balanced equation:

$$\ce{
CH3COOH~(aq) + H2O(~l) &lt;=&gt; CH3COO- ~(aq) + H3O+~(aq)
}$$

The addition of $\ce{OH-}$ would shift equilibrium to the left by Le Chatelier's principle making the newly generated $\ce{CH3COO-}$ react with $\ce{H3O+}$ to make more $\ce{CH3COOH}$, thereby increasing pH. After that, the $[\ce{H3O+}]$ is so low that the addition of the anion is not enough to keep shifting the equilibrium position to the left, so the remaining weak acid resists the changes in pH with its dissociation.

Does this make any sense?
Unfortunately this is not correct. 


  $$\ce{CH3COOH~(aq) + H2O(~l) &lt;=&gt; CH3COO- ~(aq) + H3O+~(aq)}$$


If you have a pure acetic acid solution the equation you state governs the pH of the solution.

If you are now adding a strong base like $\ce{{}^{-}OH}$, you are removing the hydronium ion from the right hand side of the equation. According to Le Châtelier, the equilibrium adapts and shifts more to the right, i.e. more acetic acid dissociates. As you remove $\ce{H3O+}$ the pH increases.
Another way of looking at this is, that you remove acetic acid from the left hand side (similar to what you stated).
$$\ce{H3CCOOH + {}^{-}OH &lt;=&gt; H3CCOO- + H2O}$$
According to Le Châtelier, the equilibrium (in the first equation) has to shift to the left, to regenerate more acetic acid, removing once again $\ce{H3O+}$ and the pH increases. 

In any case you choose to look at, you increase the concentration of acetate in the solution and another equilibrium comes into play.
$$\ce{H3CCOO- + H2O &lt;=&gt; H3CCOOH + {}^{-}OH}$$
This equilibrium is essentially responsible for raising the pH.