If I rode a space elevator to the top, assuming only natural forces apply, would the Earth be over my head or under my feet? If it was over my head how is that possible?
Above geosynchronous orbit, so called centrifugal force exceeds force of gravity. So earth would be above you.
Centrifugal acceleration (actually just inertia in a rotating frame) is $\omega^2r$ where ω is angular velocity in radians.
Earth's sidereal day is about 23.93 hours, so ω is about 2 pi radians/23.93 hours. Or 7.3e-5 radians/second.
Acceleration from gravity is $GM_e/r^2$ where $M_e$ is mass of earth and r is distance from earth's center.
You will find that $GM_e/r^2$ and $\omega^2r$ exactly cancel at geosynchronous altitude. Above geosynchronous $\omega^2r$ exceeds $GM_e/r^2$. You will weigh more as you move further beyond geosynchronous height.
If my arithmetic is right, you wouldn't feel a full g until you're about 1.8 million kilometers from earth's center.
Everywhere on the Space Elevator, "up" is the direction toward the geosynchronous point (GEO). Below GEO, gravity is exerting a stronger force on your body than the centrifugal force caused by rotation -- "down" is toward Earth; beyond GEO, the centrifugal force is stronger than gravity, so the net force on your body is away from the Earth. Right at GEO there's no "up" or "down," and indeed for some distance on either side of GEO the net force is too small for your senses to detect.
From the surface you climb up to GEO and then, without changing direction, you climb down to the counterweight at the outer end of the elevator.
The rule for "what holds the Space Elevator up" is that the sum of the net force on all the mass of the elevator above GEO must be larger than the sum of the net force on all the mass below GEO.