The gravitational potential is given as $$U(r)=-\frac
What is the significance of a reference point in calculating the potential?
It is merely a convenience. You can add an arbitrary constant to the energy without changing the physics. The reference point is usually chosen to make some common subset of problems easier.
The gravitational potential is given as $U(r)=-\frac$ . and R is the distance of the object from the Center of the earth . and if I’m calculating the potential near the surface of the earth it is mgh. mgh is obtained by considering the surface of the earth as zero potential.
Supposing that $R_e$ is the radius of the earth, you can write: $$ R=R_e+h\;, $$ where $h$ is small compared with $R_e$ .
You can thus perform a Taylor series expansion in the small parameter $h/R_e$ (small compared to $1$ ): $$ U=-\frac\frac \approx -\frac\left(1 - \frac\right) =C + \frach \equiv C + mgh\;, $$ where $C$ is a constant and $g\equiv \frac$ .
As you mention, you can always redefine the zero of the potential energy by subtracting off the constant $C$ : $$ \tilde U(h) = U - C = mgh $$