Potential Energy in conservation of energy In energy conservation theorem, everywhere $PE=mgh$ is used. What I found is that change in PE= mgh.
According to me, while calculating total energy, instead of change in PE, total PE should be used which is ofcourse in case of only gravitational field, that is $\frac {GMm}{R}$.
Then, why do we use change in PE?
 A: That is make things only more complex. Yes it is true for conservation of energy we should use total PE but still it will not matter as in conserving energy also, we are dealing with change in energy. We can say, in conservation of mechanical energy,:
Total KE +total PE =Constant
So,   change in KE + change in PE =0. 
the final result doesn't matter. 
Other way you can understand is that for this case,we choose the reference of PE at the ground or PE(ground)=0. So the expression for PE  changes to mgh. 
A: The formula $PE = mgh$ applies ONLY for very small changes in height relative to the radius of the earth (e.g., a few thousand meters or less), and is typically used close to the surface of the earth.  In addition, when using this formula, the choice for $PE=0$ is arbitrary, meaning that you can place that point anywhere you want to place it.  Typically, this point is selected in the most convenient place, with the goal of making the associated algebra as easy as practical.
For example, if I am at the surface of the earth, and I want to measure the gravitational potential energy of an object that I am elevating, I would typically choose $PE=0$ at the surface of the earth, with all points above that being associated with positive gravitational potential energy.  However, if I know that the maximum elevation will be 10 meters, I could just as easily choose $PE=0$ at a height of 10 meters, and all points below that mark would be associated with negative gravitational potential energy.  Regardless of which point I choose as $PE=0$, all increases in height will be associated with a positive change in gravitational potential energy and all decreases in height will be associated with a negative change in gravitational potential energy.  Thus, as gravitational potential energy is gained, kinetic energy is lost, and vice versa, and conservation of energy still holds, regardless of where I place the point at which $PE=0$.
A: Potential energy as well as kinetic energy depends on the frame of reference with respect to which it is measured.
A box of mass $m$ sits on the surface of a table a height $h$ above the floor. With respect to the surface of the table, it has zero gravitational potential energy. With respect to the floor, its potential energy is $mgh$. If the room happens to be on the second story of a building, it has a potential energy with respect to the earth of is $mg$ times the height of from the surface of the earth to the surface of the table. 
Similarly, kinetic energy of a mass $m$ is $\frac{mv^2}{2}$. It depends on the reference frame with respect to which the velocity $v$ is measured. An automobile has mass $m$ and velocity $v$ with respect to the frame of reference of a person standing on the side of the road. With respect to the person on the road its kinetic energy is $\frac{mv^2}{2}$.  With respect to frame of reference of the driver it is zero. With respect to another car moving toward the the first with the same velocity, each car has kinetic energy of $2mv^2$ with respect to the other. You get the idea. 
On the other hand, changes in potential and kinetic energy are not reference frame dependent. If the box on our table is lifted a height $y$ above the surface of the table, the change in potential energy is $mgy$ with respect to the surface of the table, with respect to the surface of the floor of the room, and with respect to the surface of the ground outside the building. The same applies to changes in kinetic energy. 
Regardless of the frame of reference, $\Delta KE +\Delta PE=0$ for conservation of energy. 
Hope this helps.
A: Okay so here is a short proof on how the formula $U_r-U_{\infty}=-GMm/r$ is derived.
As you should see, the potential energy at $r=\infty$ is considered to be zero. This is done because for defining any potential, you need to have a reference ie a point where potential can considered to be zero. So talking about absolute potential at any point is not possible.
While using the formula $U = mgh$, your reference point changes. You now consider the potential to be zero at the surface of the earth.

Significance of Negative Sign

Well that's just a sign convention. What you should note is that generally the sign is negative when the force acting is attractive in nature and vice versa.
