Dissociated and non dissociated acetic acid react with $\ce{KOH}$ in the same way. This result does not depend on the concentrations of the initial acid and of the initial hydroxide. It will produce potassium acetate.
Now it is also right that the reaction is not total, as potassium acetate is slightly hydrolized. But this effect is rather weak. A solution $1$ molar in potassium acetate has a pH $9.37$. This corresponds to a residual $\ce{OH-}$ concentration equal to $\ce{2.35·10^{-5}}$ mole par liter.
So out of a solution containing $1$ mole potassium acetate per liter, only $23.5$ micromole are decomposed into acid and hydroxide. This ins not zero of course, but it is rather small. $23.5$ parts per million are not transformed into potassium acetate. The purity of the obtained potassium acetate is $100$% - $0.00235$% = $99.99765$ %
As many comments have been written about my very first sentence, I think advisable to develop it in some detail. An acetic acid aqueous solution is made of a vast majority of unchanged $\ce{CH3COOH}$ molecules, plus a tiny percentage of ions due to the reaction
$$\ce{CH3COOH <=> CH3COO- + H+ \tag{1}}$$ which could also be written $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+ \tag{2} }$$
This reaction is an equilibrium, with an equilibrium constant which is usually derived from equation $(1)$ : $K\pu{_a = \frac{[H^+][CH_3COO-]}{[CH_3COOH]} = 1.8 10^{-5} M}$. In a $1$ molar $\ce{CH3COOH}$ solution, the concentration $\ce{[H^+]}$ is $\sqrt{K\pu{_a}} = 4.24· 10^{-3}$ M. Whatever its numerical value, it means that this concentration $\ce{[H+]}$ is rather small, with respect to $1$ M.
However, if these rare $\ce{H+}$ ions are suddenly destroyed by adding some drops of a $\ce{KOH}$ solution, the concentration $\ce{[H+]}$ should fall to zero. It does not, as immediately Nature reacts, and enough $\ce{CH3COOH}$ molecules get dissociated according to ($1$) or ($2$) to compensate for the sudden lost of $\ce{H+}$. As a consequence, the concentration of acetic acid decreases and the concentration of acetate ions increases. Apparently this reaction ($1$) or ($2$) is extremely fast, as it is impossible to follow its kinetics by spectrometric methods. Why is this reaction so fast ? It is hard to explain, because it is a reaction from or between neutral substances. And usually reactions from or between neutral substances are not immediate, and their kinetics can be followed by spectrometric methods.
On the contrary, every chemist knows that reactions due to ions (like neutralizations) are so fast that their kinetics cannot be determined. This is why some chemists have emitted the idea that the added ions $\ce{OH-}$ should react directly with the most abundant substance in the solution, namely the molecule $\ce{CH3COOH}$ in a possible reaction $\ce{OH- + CH3COOH -> CH3COO- + H2O}$. The $\ce{H+}$ ions would not wait for the equilibrium to be re-adjusted.
But I would not fight for this idea, that I have read once in a manual. It is not the point of the present discussion. I would not abject if the reader prefers thinking that the added $\ce{OH-}$ react with the ions $\ce{H+}$ (or $\ce{H3O+}$) produced by the quick adjustment of reactions ($1$) or ($2$).