Is Gravitational Force (using $F=Gm_1m_2/r^2$) only valid for point masses? If one of the bodies between which there is a gravitational force, is not a point mass, can we use Newton's Univeral Law of Gravitation Equation $F = GMm/r^2$?
Is this equation still valid in this case?
Or do we have to use Gravitational Field Equation: $g$ = $F$(Force of Attraction)/$m$ (Small Test Mass), instead of F = GMm/r^2?
So $g = GM/r^2$ ($GM/r_1 r_2$); ($F$ will be $GMm/r_1 r_2$).
or is inverse square law not accurate for non-point masses as in $Gm_1m_2/r^2$ (where $f ∝ 1/r^2$)?
 A: It sort of depends on how you use $F = GMm/r^2$ and your semantics. A more in-depth explanation is required here.
A naive application of this formula on extended bodies would be to somehow assign to each extended body a single point/location and then apply the formula. (Remember that if a body is extended then it doesn't have a single point location.) This naive approach only works in limited circumstances: it works in the case where both bodies are really far away from each other so that the gravitational field strength from one body has negligible variation over the volume of the other body, and vice-versa. If this condition is not met, then our naive approach fails for at least two reasons:

*

*First, we might expect that we can use the center of masses to assign positions to extended masses. This is false, because the center of mass will not coincide with the center of gravity, so you need to use the center of gravity. However, to derive the center of gravity you need to step away from the naive view of $F = GMm/r^2$ and use the approach below.

*Second, the extended body will have tidal forces acting upon it, and these simply can't be accounted for in our naive approach. I don't talk about tidal forces below, because I will focus only on total forces on bodies, but it is still something worth pointing out. If you want to think about tidal forces, then you simply can't substitute extended bodies with point particles.

A non-naive application of $F = GMm/r^2$ would be to take the extended bodies, imagine them being subdivided into smaller and smaller masses, and then somehow "sum" the forces between the little masses to get a total force $F$. The limit of taking smaller and smaller masses will result in a more complicated formula involving integrals. The formula is
$$ \vec{F}_{A\text{ on }B} = \int_{A}\int_{B} -G\frac{dm_{A}dm_{B}}{|\vec{r}_{1} - \vec{r}_{2}|^{2}}\hat{e}_{r} $$
where the integrals are taken over the masses of bodies $A$ and $B$, $\vec{r}_{1}$ is the position of mass element $dm_{A}$, $\vec{r}_{2}$ is the position of mass element $dm_{B}$, and $\hat{e}_{r}$ is the unit vector in direction $\vec{r} = \vec{r}_{2} - \vec{r}_{1}$.
Technically or literally speaking, this formula is different from the original $F = GMm/r^2$. We postulate this as a generalization of the original formula. However, one can also say we are still using $F = GMm/r^2$ because the integral formula comes from our physical intuition of subdividing extended masses into smaller and smaller masses and applying the formula for point particles. I think the question of whether you are still using the original formula becomes a matter of semantics once you understand what is going on.

Field Approach
The field approach is helpful in taking things apart and digesting what is going on, but it is not absolutely necessary. We can in principle just use all the stuff mentioned above. Nonetheless I will try to go through some steps to show how we can generalize from point-masses to extended bodies using the field approach.
We initially postulate that Newton's law applies to point-masses only. Given point-masses $m_{A}$ and $m_{B}$ with separation distance $r$, the gravitational force of $A$ acting on $B$ is
$$ \vec{F}_{A\text{ on }B} = -G\frac{m_{A}m_{B}}{|\vec{r}_{A} - \vec{r}_{B}|^{2}}\hat{e}_{AB} $$
where $\vec{r}_{A}$ is the location of point $A$, $\vec{r}_{B}$ is the location of point $B$, and $\hat{e}_{AB}$ is the unit vector pointing from point $A$ to point $B$. The negative sign in the above expression means the force vector points from point $B$ to point $A$, so gravity causes $B$ to move into the direction of $A$, as expected.
Of course, we want to generalize this to realistic extended bodies. One way to do this is the way you suggested, which is to consider gravity as a gravitational field $\vec{g}$. The gravitational field due to a point-mass $m_{A}$ is
$$ \vec{g}(\vec{r}) = -G\frac{m_{A}}{|\vec{r} - \vec{r}_{A}|^{2}}\hat{e}_{r_{A}r} = -G\frac{m_{A}}{|\vec{r} - \vec{r}_{A}|^{3}}(\vec{r} - \vec{r}_{A}) $$
where $\vec{r}_{A}$ is the location of $A$ and $\hat{e}_{r_{A}r}$ is the unit vector pointing from $\vec{r}_{A}$ to $\vec{r}$.
Now a natural way to generalize this to extended bodies is to postulate that, given an extended body $A$, the gravitation field due to $A$ is
$$ \vec{g}(\vec{r}) = \int_{A} -G\frac{dm_{A}}{|\vec{r} - \vec{r}\,'|^{3}}(\vec{r} - \vec{r}\,') = \int_{A} -G\frac{\rho(\vec{r}\,')\, d^{3}r'}{|\vec{r} - \vec{r}\,'|^{3}}(\vec{r} - \vec{r}\,') $$
where the right-most side has a volume integral over the volume of body $A$ and $\rho(\vec{r}\,')$ is the density of $A$ at $\vec{r}\,'$. Then the force of this gravitational field on a point test mass $m_{B}$ is
$$ \vec{F}_{A\text{ on }B} = m_{B}\cdot\vec{g}(\vec{r}_{B}) $$
where $\vec{r}_{B}$ is the location of point $B$.
We can further use this to consider the force by $\vec{g}$ on an extended body $B$. This would be
$$ \vec{F} = \int_{B} dm_{B}\cdot\vec{g}(\vec{r}\,') = \int_{B} \rho(\vec{r}\,')\cdot \vec{g}(\vec{r}\,')\, d^{3}r' $$
where the right-most side is a volume integral over $B$ and $\rho(\vec{r}\,')$ is the density of $B$ at $\vec{r}\,'$.
At this stage, we can find the gravitational field $\vec{g}$ due to an extended body $A$, and then use this to calculate the force of gravity by $\vec{g}$ on an extended body $B$. Thus, we have completed our generalization.
Now this formulation has many analogues to the formulation of electrostatics, and so there are various theorems of Newtonian gravity that are analogous to theorems of electrostatics. One of these is Gauss's law for gravity. When you apply Gauss's law for gravity to a uniform spherical mass, you find out that the gravitational field is exactly the same as that of a point-mass located at the center of the spherical mass.
Note that we even though we can substitute a uniform spherical body with a point-mass at the center if we want to find $\vec{g}$ due to that body, we can't do this if we want to look at the effect of $\vec{g}$ on that body. Instead, we'd need to use the center of gravity, which is not necessarily at the center of the sphere.
Actually, what is interesting is that the corollary to Gauss's law mentioned above is called the shell theorem, and it was derived by Isaac Newton before electrostatics and before the field approach was ever born.
A: To extend Newton's laws to macroscopic bodies, one considers anything that is not a punctual mass as a set of punctual masses and apply Newton's laws between them.
This method gives rises to macroscopic mechanics that encompasses rigid body mechanics, elasticity and fluid dynamics.
A: About the briefest answer is that the Newtonian gravitational force equation $F = G.m_1.m_2/r^2$ is validly applicable to a point mass, and also to a spherically-symmetrical mass-distribution for the force at points outside the mass distribution (reckoned in the equation as if represented by its center).
(In the latter situation, the applicability works both ways: either or both of the masses taking part in the interaction described by the equation may be spherically symmetrical mass-distributions.)
(In Newton's 'Principia' this is shown at Book 1 sec.12, starting at Props. 70 and 71, with various corollaries developed in Props. 72-84. More modern re-derivations have of course been developed since then, as shown in Maximal Ideal's answer and other work referring to Newton's shell theorem.)
(At points within a spherical shell of uniform surface mass-density, the cited proofs show that the shell has zero Newtonian gravitational attraction.  So for points within such a mass-distribution, at distance $r_i$ from the center, the gravitational force equation can conditionally still be applied if its considered mass is reduced by leaving out those parts of the mass located at distances from the center greater than $r_i$. Thus, to the extent that the Earth could be approximated as spherically symmetrical and of uniform density, the gravitational acceleration $g$ would decrease linearly from the surface down to zero at the center.)
The expression "center of gravity" (for the center of mass or centroid) is a trap in this context. Historically it arose before discovery of the law of gravitation; it does appear to describe the center of heaviness around which an extended mass can be suspended or supported in balance.  But in spite of its being pointed out in many places that the "center of gravity" is not a center of gravitational attraction, one can find (especially online) some effectively fictional expositions that depend on the mistaken assumption sometimes left implicit that a center of gravity is also a center of attraction, even in highly asymmetrical cases of mass-combination.
