Are these corrections are distinct from the Earth's rotation (or do they express the same thing really)? Can they be compensated by the Earth's rotation?

Yes and no. Those are different effects.

As a clear example, consider tides of our oceans. At one place on the Earth's equator the tides are rising, and at another they are falling. In rotating frame the centrifugal force is the same along whole equator, so there is no way to attribute this effect to rotation.

In Luboš answer he claims, that center of the Earth is moving along the geodesic (i.e. is inertial), but the further away you are from the center, the bigger noninertial effects you will experience. This effects are due to the fact, that two parallel geodesics will not remain parallel for long - there is a geodesic deviation. So if we have two freely falling particles in the vicinity of Earth that are initially parallel (in four-dimensional view, 3D parallelism does not suffice), they will start to move away from/closer to each other, just like in figure 4 in this wiki page. The formula for tidal effects is given in the page, together with some values.

But the thing is, we are mixing two frameworks together. In GR, one could hardly consider Earth inertial frame, since everything on the surface of the Earth is accelerated upward with pretty significant acceleration. Every engineer accounts for this when he constructs car or building, so the only way in which to interpret the question whether Earth is inertial frame is in the framework of Newtonian gravity. There, the tidal forces are not due to the failure of our frame to be inertial, but due to the gravitational interaction between bodies. From this point of view, the Earth can be considered inertial.

But we know, it rotates around its own axis and around Sun. Rotation around axis makes surface noninertial as you correctly stated in your answer. If we fix the rotation of our frame by the distant stars, then we get rid of this and rotation around Sun will produce the larges noninertiality in our frame.

In Newtonian framework, one does not consider freely falling bodies inertial, but accelerated. However, from equivalence principle we know, that this acceleration of freely falling bodies simulates inertial frame pretty accurately, we just need to ignore gravitational force that governs the free fall. In Newtonian framework, to describe motion of Earth around Sun in the frame of the Earth, we would need to compute gravitational force from the Sun and centrifugal force from noninertiality of our frame and we arrive at conclusion that they cancel each other out. This is mathematically equivalent to asserting, that frame of Earth in vicinity of it is inertial and there is no gravitational field from the Sun. This is in fact what engineers do. They assume Earth is inertial frame and they do not bother with Sun's gravity. So the approximation that uses Earth's frame as inertial frame is much better than what we would have guessed just by looking at the motion of the Earth and seeing that it moves around an ellipse.

That being said, Earth's center does not move on a geodesic, but this is different effect than tidal forces that Luboš describes. This arise due to the fact, that each particle constituting the Earth tries to move on its own geodesic, but intermolecular forces prohibit this and resulting motion one could call a compromise. But this is very small effect, probably not even worth mentioning.

Are these corrections are distinct from the Earth's rotation (or do they express the same thing really)? Can they be compensated by the Earth's rotation?

The are distinct and they cannot.

As a clear example, consider tides of our oceans. At one place on the Earth's equator the tides are rising, and at another they are falling. In rotating frame the centrifugal force is the same along whole equator, so there is no way to attribute this effect to rotation.

In Luboš answer he claims, that center of the Earth is moving along the geodesic (i.e. is inertial), but the further away you are from the center, the bigger noninertial effects you will experience. This effects are due to the fact, that two parallel geodesics will not remain parallel for long - there is a geodesic deviation. So if we have two freely falling particles in the vicinity of Earth that are initially parallel (in four-dimensional view, 3D parallelism does not suffice), they will start to move away from/closer to each other, just like in figure 4 in this wiki page. The formula for tidal effects is given in the page, together with some values.

But the thing is, we are mixing two frameworks together. In GR, one could hardly consider Earth inertial frame, since everything on the surface of the Earth is accelerated upward with pretty significant acceleration. Every engineer accounts for this when he constructs car or building, so the only way in which to interpret the question whether Earth is inertial frame is in the framework of Newtonian gravity. There, the tidal forces are not due to the failure of our frame to be inertial, but due to the gravitational interaction between bodies. From this point of view, the Earth can be considered inertial.

But we know, it rotates around its own axis and around Sun. Rotation around axis makes surface noninertial as you correctly stated in your answer. If we fix the rotation of our frame by the distant stars, then we get rid of this and rotation around Sun will produce the larges noninertiality in our frame.

In Newtonian framework, one does not consider freely falling bodies inertial, but accelerated. However, from equivalence principle we know, that this acceleration of freely falling bodies simulates inertial frame pretty accurately, we just need to ignore gravitational force that governs the free fall. In Newtonian framework, to describe motion of Earth around Sun in the frame of the Earth, we would need to compute gravitational force from the Sun and centrifugal force from noninertiality of our frame and we arrive at conclusion that they cancel each other out. This is mathematically equivalent to asserting, that frame of Earth in vicinity of it is inertial and there is no gravitational field from the Sun. This is in fact what engineers do. They assume Earth is inertial frame and they do not bother with Sun's gravity. So the approximation that uses Earth's frame as inertial frame is much better than what we would have guessed just by looking at the motion of the Earth and seeing that it moves around an ellipse.

That being said, Earth's center does not move on a geodesic, but this is different effect than tidal forces that Luboš describes. This arise due to the fact, that each particle constituting the Earth tries to move on its own geodesic, but intermolecular forces prohibit this and resulting motion one could call a compromise. But this is very small effect, probably not even worth mentioning.

Are these corrections are distinct from the Earth's rotation (or do they express the same thing really)? Can they be compensated by the Earth's rotation?

Yes and no. Those are different effects.

As a clear example, consider tides of our oceans. At one place on the Earth's equator the tides are rising, and at another they are falling. In rotating frame the centrifugal force is the same along whole equator, so there is no way to attribute this effect to rotation.

In Luboš answer he claims, that center of the Earth is moving along the geodesic (i.e. is inertial), but the further away you are from the center, the bigger noninertial effects you will experience. This effects are due to the fact, that two parallel geodesics will not remain parallel for long - there is a geodesic deviation. So if we have two freely falling particles in the vicinity of Earth that are initially parallel, they will start to move away from/closer to each other, just like in figure 4 in this wiki page. The formula for tidal effects is given in the page, together with some values.

There are also other effects on top of tidal forces and rotation. One is the fact, that Earth is not perfectly spherical and thus Birkhoff's theorem does not really apply. Another, smaller one, is the fact that Earth's center is not really moving along the geodesic. Each particle that constitutes Earth tries to move on its own geodesic, but intermolecular forces prohibit this and resulting motion one could call a compromise. These are however much smaller than the tidal forces of Sun and Moon.

There are also effects due to self-gravity of Earth. The traditional principle of equivalence works for test particles, i.e. particles whose motion is not influenced by their own gravity, but at certain approximation this assumption also breaks down.

Are these corrections are distinct from the Earth's rotation (or do they express the same thing really)? Can they be compensated by the Earth's rotation?

Yes and no. Those are different effects.

As a clear example, consider tides of our oceans. At one place on the Earth's equator the tides are rising, and at another they are falling. In rotating frame the centrifugal force is the same along whole equator, so there is no way to attribute this effect to rotation.

In Luboš answer he claims, that center of the Earth is moving along the geodesic (i.e. is inertial), but the further away you are from the center, the bigger noninertial effects you will experience. This effects are due to the fact, that two parallel geodesics will not remain parallel for long - there is a geodesic deviation. So if we have two freely falling particles in the vicinity of Earth that are initially parallel (in four-dimensional view, 3D parallelism does not suffice), they will start to move away from/closer to each other, just like in figure 4 in this wiki page. The formula for tidal effects is given in the page, together with some values.

But the thing is, we are mixing two frameworks together. In GR, one could hardly consider Earth inertial frame, since everything on the surface of the Earth is accelerated upward with pretty significant acceleration. Every engineer accounts for this when he constructs car or building, so the only way in which to interpret the question whether Earth is inertial frame is in the framework of Newtonian gravity. There, the tidal forces are not due to the failure of our frame to be inertial, but due to the gravitational interaction between bodies. From this point of view, the Earth can be considered inertial.

But we know, it rotates around its own axis and around Sun. Rotation around axis makes surface noninertial as you correctly stated in your answer. If we fix the rotation of our frame by the distant stars, then we get rid of this and rotation around Sun will produce the larges noninertiality in our frame.

In Newtonian framework, one does not consider freely falling bodies inertial, but accelerated. However, from equivalence principle we know, that this acceleration of freely falling bodies simulates inertial frame pretty accurately, we just need to ignore gravitational force that governs the free fall. In Newtonian framework, to describe motion of Earth around Sun in the frame of the Earth, we would need to compute gravitational force from the Sun and centrifugal force from noninertiality of our frame and we arrive at conclusion that they cancel each other out. This is mathematically equivalent to asserting, that frame of Earth in vicinity of it is inertial and there is no gravitational field from the Sun. This is in fact what engineers do. They assume Earth is inertial frame and they do not bother with Sun's gravity. So the approximation that uses Earth's frame as inertial frame is much better than what we would have guessed just by looking at the motion of the Earth and seeing that it moves around an ellipse.

That being said, Earth's center does not move on a geodesic, but this is different effect than tidal forces that Luboš describes. This arise due to the fact, that each particle constituting the Earth tries to move on its own geodesic, but intermolecular forces prohibit this and resulting motion one could call a compromise. But this is very small effect, probably not even worth mentioning.