Suppose I have a principle vector bundle $\pi:E\rightarrow M$ with a group $G$. By definition, the action of $G$ preserves fibers, so $g\circ\pi=\pi\circ g$. This means that the differential maps satisfy $(dg)_{\pi(e)}\circ(d\pi)_e=(d\pi)_{ge}\circ (dg)_e$. Any vector $v$ in the vertical bundle of $E$ will be annihilated by the left-hand side; since $g$ is invertible, then $dg$ is invertible, so $(dg)_e(v)$ must be be in the kernel of $(d\pi)_{ge}$. Thus, $(dg)_eV_e\subseteq V_{ge}$. Again since $dg$ is invertible, we have equality.
This matches the definition of a "principal Ehresmann connection". So it seems like with horizontal bundles, we must enforce this property, but it happens automatically for vertical bundles. Is this correct?
Up to the error pointed out in the remark of @Berci you are right. The vertical subbundle of a prinicpal bundle is invariant under the principal right action. However, I think your last sentence is kind of misleading. The point is that the vertical subbundle is canonical, and it is preserved by (the derivative of) any bundle map. By assumption, the principal action is by bundle maps (covering the identity) so it preserves the vertical subbundle. The horizontal distribution provided by a principal connection represents a choice and except for smooth dependence on the point, there is no relation between the choices in different points of a fiber is required. Compatibilty with the principal action is a natural additional condition to require, but there is no reason why it should be satisfied automatically.