In Layman’s views, you can also think of it this way: initially, you have water, a small amount of $\ce{H3O+}$ and a small amount of $\ce{OH-}$. If you add an external proton source, we can consider what will happen to each of these:
\begin{array}{lccc}\hline
\text{time point} & \text{acid} & \text{neutral} & \text{base} \\ \hline
\text{before a.} & \ce{H3O+} & \ce{H2O} & \ce{OH-} \\
\text{after a.} & \text{no change} & \ce{H3O+} & \ce{H2O} \\ \hline \end{array}
In words: Nothing will happen to $\ce{H3O+}$ — at least nothing notable; $\ce{H4O^2+}$ is too unstable in aquaeous solution to contribute meaningly. Water will be protonated to $\ce{H3O+}$ and hydroxide ions will also be protonated (statistically) by the incoming acid to give $\ce{H2O}$. Thus, you can consider the addition of acid to both increase oxonium ion concentration and reduce hydroxide ion concentration.
The equilibrium constant and equation is just a more formal and scientifically more correct way of putting this.