$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$
My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?
You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$\ce{H3O+ + OH- <=> 2 H2O}$$
If you add up the two reaction, you get a third one:
$$\ce{CH3COOH + OH- <=> H2O + CH3COO-}$$
You can go either way (use the second or the third reaction) to explain that $\ce{OH-}$ is part of a neutralization reaction. As a consequence, the concentration of $\ce{CH3COO-}$ increases and that of $\ce{CH3COOH}$ and hydronium decreases.
Hydroxide reacts with hydronium
In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $\pu{e-5}$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.
Hydroxide reacts with acetic acid
In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.
Which way is better?
It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.
Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?
No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.