Why doesn't planet Earth expand if I accelerate upwards when standing on its surface?

Because of Earth's gravity (and rotations, but we will focus on gravity since that seems to be the point of your question) you are in an accelerated frame of reference, not an inertial one. Near the surface of Earth g is about 9.8 meters per second squared. This will make your weight be about the same on Earth as it would seem to be if you were accelerating at 9.8 meters per second squared out in space, far enough away from any other body for gravity to be negligible.

Spacetime curvature makes this possible. Here's an analogy. There are two paths on opposite sides of the equator, at a constant distance from it. Someone walking east along the path north of the equator will have to continually turn slightly left to stay on the path. (If that isn't obvious, imagine it's so far north that it visibly circles the pole.) Likewise, someone walking east on the path south of the equator will have to turn right. Two people walking side by side along the paths will stay the same distance apart, even though they're constantly turning away from each other. This wouldn't be possible on the Euclidean plane, but it's possible on a curved surface. That's what happens in general relativity, but the direction they're walking is the time direction, and the turning is acceleration.

It is accelerated from YOUR referential. In free fall you will follow the space-time geodesics. But the earth's ground prevents you from falling toward the center of mass of the earth. So in your referential, you are accelerated upward by the ground.

It comes down to a definition of acceleration. Acceleration is most universally appreciable as a force application contradicting an object's natural position or trajectory. Notice that this does not require that the object move -- only that it is being affected by a force, as in 'experiencing pressure'.

So by this definition an object at apparent rest on a table top is being forced by the solid table surface, and feels the pressure of this force throughout its form, and so on.

It also helps to appreciate gravity as an electromagnetic phenomenon, as the definition of acceleration also applies to (ferro-)magnetic forces. When you see two strong magnets pulling or pushing each other, it appears that they are exerting a force, as though expending energy...

But to the magnets their unhindered relative motion represents a state of rest given their natural atomic states. Energy expense is only experienced by the person holding the magnets apart/together against the natural tendency, and by the magnets when they are prevented from their natural relative motion (including if/when they impact).

Disregarding the earth rotation for being too slow, we can use the Schwarzschild metric as a good approximation:

$$c^2dtau^2 = left(1 - frac{2GM}{c^2r}right)c^2dt^2 - frac{1}{left(1 - frac{2GM}{c^2r}right)}dr^2 - r^2dtheta^2 - r^2sin^2theta dphi^2$$

For a body at rest on the earth surface, $$dr = 0,, dtheta = dphi = 0impliesleft(frac{dt}{dtau}right)^2 = frac{1}{left(1 - frac{2GM}{c^2r}right)}$$

The second covariant derivative of $r$ with respect to $tau$, is:

$$nabla^2_{tau}r = frac{partial^2 r}{partial tau^2} - sum{Gamma_{ij}^rfrac{partial X^i}{partial tau}frac{partial X^j}{partial tau}}$$

Most of the terms of the summation are zero because the body is at rest. The first term of the right side is the conventional acceleration, that is also zero for the same reason. The non zero terms are:

$$nabla^2_{tau}r = left(1 - frac{2GM}{c^2r}right)left(frac{GM}{r^2}right)left(frac{partial t}{partial tau}right)^2 = frac{GM}{r^2}$$

That is our $g$.

That is because we are just accelerating radially outwards and not "moving" radially outwards. This case is analogous to circular motion where there is radial acceleration but no radial movement.

You can refer to my article in the link below for detailed explanation: https://paribeshregmi.medium.com/a-soft-intro-to-general-relativity-aa46da221747