What is the physical meaning and origin of the gravitational constant $G$? Is $G$ just a value to fix the units in the equation? Something like a proportionality constant or a coupling constant?:
$$F=\frac{G m_{1} m_{2}}{r^{2}}$$
Does it have any physical meaning or physical origin?
What does it represent in General Relativity theory?
 A: 
Is G just a value to fix the units in the equation? Something like a proportionality constant or a coupling constant?

As you say, it is just a value to fix the units in the equation. There is no physical content to its value other than to describe the size of the unit of mass compared to the units of length and time.

What does it represent in General Relativity theory?

Simply a unit conversion. Usually we simply get rid of it by using units such that $c=G=1$. These are so commonly used that they are called "natural units". These include Planck units and Geometrized units among others.

Does it have any physical meaning or physical origin?

No, or rather no physical meaning beyond the choice of units. The physically meaningful constants are dimensionless. To get physically measurable differences requires a change to one of the dimensionless constants.
A: (G) first appeared in Newton's equation and is needed to predict the magnitude of the gravitational force between two known masses. (It's value will depend on the system of units you are working with.) Its value was first determined with the Cavendish apparatus:  a spherical mass on each end of a thin rod, suspended on a fine wire, and brought near to two other spherical masses. I'm guessing that it is still the least accurately known of the various physical constants.
A: This answer is about the physical origin of gravity.
It comes about due to the requirement that the universe should have scaling symmetry and the principle of conservation of energy.

If the universe were to rescale, then all physical constants and quantities with length dimensions would vary as in the table below, where $H$ is a scaling constant.
\begin{array}{c|c|c}
{quantity} & {length-dimension} & {change}\\
\hline
length & 1 & e^{Ht}\\
mass & 0 & constant\\
time & 0 & constant\\
h & 2 & e^{2Ht}\\
c & 1 & e^{Ht}\\
G & 3 & e^{3Ht}\\
Area & 2 & e^{2Ht}\\
\end{array}
Then imagine a mass $m$ of energy $mc^2$, it's energy changes to $mc^2e^{2Ht}$.
That's without gravity, however with gravity the total energy due to the mass is
$$mc^2 - \frac{GMm}{R}$$
Where $M$and $R$ represent the mass and radius of the universe up to the Hubble radius and small numerical constants are omitted for simplicity.
Then energy can be conserved as the universe rescales if
$$(mc^2 - \frac{GMm}{R})e^{2Ht} = 0$$
and $$G=\frac{Rc^2}{M}$$
That gives a reason for the value of $G$ and an explanation of the flatness problem.
The equivalent from General Relativity is
$$G=\frac{3H^2}{8\pi \rho}$$
i.e the universe is at critical density.
See also this link
