I understand that in a non-inertial frame of reference rotating with the Earth, an object at rest has a weight that is equal to mg. This mg is the vector addition of the centrifugal force and the true gravitational vector.
This part is correct. This is how g is defined.
Thus mg is simply the addition of the vertical components of the true gravitational vector and the vertical component of the centrifugal force as the horizontal components cancel out (object at rest).
This part is incorrect. The centrifugal force is a very small part of g. It is greatest at the equator, and even there, it is only about 0.0035 g. The centrifugal force does not cancel out the gravitational force.
You are ignoring that the ground is pushing up on you. In a frame rotating with the Earth, the correct expression for a person standing at rest on the surface of the Earth is $m\vec g + \vec N = 0$.
What about an inertial frame? The only forces acting on that person from the perspective of the inertial frame are the true forces, Newtonian gravitation and the normal force. Given that $m\vec g$ is Newtonian gravitation plus the centrifugal force, it's easy to calculate the Newtonian gravitational force acting on a person. The centrifugal force acting on the person is $-m\, \vec\omega \times (\vec\omega \times \vec r)$. The Newtonian gravitational force is thus $m\left(\vec g + \vec\omega \times (\vec\omega \times \vec r)\right)$. The net true force acting on the person is thus
$$m\left(\vec g + \vec\omega \times (\vec\omega \times \vec r)\right) + \vec N = m\left(\vec g + \vec\omega \times (\vec\omega \times \vec r) - g\right) = m\,\vec\omega \times (\vec\omega \times \vec r)$$
This is not zero. It is in fact uniform circular motion, with the force always pointing at the Earth's rotation axis, and a rotation rate of one rotation per sidereal day. This is the behavior a person standing still on the surface of the rotating undergoes from the perspective of an inertial frame.