In words, $|{k}\rangle = \sqrt{2 \omega_k} |\vec k\rangle $ tells us that the three-momentum eigenstates $|\vec k\rangle $ are related to the corresponding four-momentum eigenstates $|{k}\rangle$ by a factor of $\sqrt{2 \omega_k}$. While I understand how to derive this starting from the definitions of the creation and annihilation operators ($a(k)$ vs. $a(\vec k)$ etc.), I'm lacking any real understanding why this relationship between three-momentum eigenstates and four-momentum eigenstates should hold.
(For a reference, see e.g. Eq. 2.64 in http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf. Moreover in Eq. 2.45 it is demonstrated that $|\vec k\rangle $ is indeed a three-momentum eigenstate.)
