1) point 1: We do not need to put the $0$ point at infinity. Since it is potential energy, we can set it to $0$ anywhere we want. This is because adding a constant to potential energy does not change the force involved (since $F=-\frac{dU}{dr}$). The reason so many people choose to do this is because what we are usually interested in is a change in potential energy, rather than an absolute value of it. If we set $U = 0$ at infinity then things work out nicely. For example, in the case of gravity at large scales, $U=-\frac{Gm_1m_2}{r}$. This expression goes to $0$ as $r$ goes to infinity. Therefore we can look at $U(r)$ as a change in potential energy from infinity, and we don't need to keep track of some arbitrary constant. The change of potential energy between two points in space becomes $$\Delta U=U(r_2)-U(r_1)=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$$
If we did want to set $U=0$ somewhere else, then we always have an arbitrary constant following us around, but then it goes away when we find changes in potential energy. Let's say $U=0$ at some point $r = R_0$. Then our function of the potential energy is $U(r)=-\frac{Gm_1m_2}{r}+\frac{Gm_1m_2}{R_0}$. This is perfectly valid physically, but we get the same result as before for the change in potential energy between two points in space:$$\Delta U=U(r_2)-U(r_1)=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{R_0}+\frac{Gm_1m_2}{r_1}-\frac{Gm_1m_2}{R_0}$$$$\Delta U=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$$ Therefore, we just set $U=0$ at infinity since it is the simplest thing to do (similar to why we put potential energy to $0$ when a spring is at its resting state).
1) point 2: This is not how you find the work done to push something against gravity in general, but we can make some assumptions to get to where it seems like you need to be. In general the work done by any force is $\int \vec F\cdot d \vec r$. So if you push on the object with some force, then this is how you determine the work done by yourself if you do not know anything else. If you know that the only forces acting on the object is yours and gravity, then you can use energy conservation as another way to get the work you have done:
$$W_{tot}=\Delta K=W_{me}+W_{grav}=W_{me}-\Delta U$$ where K is the kinetic energy. If the object starts and stops at the same speed, then we can go further:$$W_{me}=\Delta U$$ And this is probably where you getting confused. In the case where we are close to the Earth, then $W_{me}=mg \Delta h$. In the case where we are farther from the Earth, $W_{me}=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$. So your issue might be from trying to apply the first (an approximation) when you really need to be applying the second.
2) I am kind of confused on the wording here, but if I understand what you are asking, you are just wondering if you can change where $U=0$ is defined. As already stated, this is perfectly fine! You just have to make sure you stay consistent. The absolute values of your potential energies can change, as long as the relative values between points remains constant.
(Side note: You can run into issues if you try to put $U=0$ at infinity if your mass distribution is nonzero at infinity as well, but since we are just looking at gravity around the Earth we should be fine)
The potential at a point is derived with calculus,you cannot multiply $\frac{GM}{r^2}$ with the difference in positions because the gravitational field varies with distance from centre of mass of the body.Change in potential energy is defined as negative of the work done by conservative force so $$-F_{conservative} \Delta x = \Delta U$$$$F_{conservative}=- \frac{dU}{dx}$$(for infinitesimal changes).You can verify this as$$\frac{d}{dr}(\frac{-GM}{r})= \frac{GM}{r^2}$$
Best Answer
Gravitational field is a vector field and is determined by negative gradient of the gravitational potential. $$\vec g=-\vec\nabla \phi$$
Frome equation above, it is obvious that $|\vec g|=|-\vec\nabla \phi|$ (magnitude of $\vec g$ is equal to magnitude of $-\vec \nabla \phi$) and we know that $|-\vec\nabla \phi|$ is a non-negative quantity.
You have made a mistake by assuming that $|\vec g|=-|\vec\nabla \phi|$