Do forces act on points or areas? In a lot of situations we are taught about forces acting on points on solid objects (torque, point particles), and in other cases (axial stresses) we consider them as acting on an 'area', is this something that is valid for all forces, or is the 'force over an area' the result of integrating a set of forces acting on all points on the surface over that area.
Are forces pressing on an object under axial stress considered to be acting on the area as a single entity or a representation of forces acting on several points in an area.
 A: Points are a mathematical convenience. They don’t exist.
An atomic force microscope operates by touching a surface with a cantilever whose tip has a radius of curvature of a few nanometers. The cantilever can be measurably displaced by a single atom on a flat substrate.  This is as close as humans can get to a “pointlike” interaction between two extended objects.
When we talk about a “pointlike” force in a more pedestrian context, we mean that pretending the force is concentrated at a point removes more mathematical problems than it causes.  The value of the pointlike approximation depends on what you are doing. If you are using your hand to open a pickle jar, your hand and the lid are roughly the same size, and it would be silly to pretend that your hand was pointlike. But if you used the same hand to open the one-ton door to a bank vault, you would predict the same motion for the heavy door whether you applied the same force with one hand, or with two hands, or with the point of a stiletto, or with your shoulder.
Your effort to move a bank vault is “pointlike” because a bank vault is “much” bigger than you are.  Students of advanced physics spend some time learning how to quantify “much.”
A: The answer is both, at least for contact forces that are localized.

Above is a contact patch at which pressure $P(x,y)$ is applied and it varies by location. The total area of the patch is the sum of all the smaller differentiable areas $A = \int {\rm d}A$
The total force must equal the sum of pressure over the area
$$ \vec{F} = \int P(x,y)\, {\rm d}\vec{A}$$
where ${\rm d}\vec{A}$ is the differential directed area inside the patch. It is a vector with a magnitude equal to the differential area ${\rm d}A$ and direction perpendicular to the contact patch.
But the contact patch also exerts a torque about the origin, and if $\vec{r}$ is the position of every particle inside the contact patch, then this torque is
$$ \vec{\tau} = \int P(x,y)\, \vec{r}\times {\rm d}\vec{A}$$
This will create a center of pressure on the patch $\vec{c}$ defined by
$$ \vec{\tau} = \vec{c} \times \vec{F} $$
As an idealization for a rigid body the above is equivalent to a single force $\vec{F}$ going through the point $\vec{c}$.
But for deformable bodies, the shape, size, and pressure distribution of the contact patch is important in finding the deflections on the body.
A: Almost always in actual physical problems, the force is an electromagnetic force and it acts on charges.
Sometimes the force is gravity and it acts on masses.
Are there point charges or point masses? I don't know for sure, but it doesn't really matter when you're trying to compute an answer.
Inverse-square forces act on uniform spheres just like they'd act on points at the centers of those spheres. So a lot of the time it's simpler to work as if masses are point masses and charges are point charges.
On the other hand, sometimes it works better to assume that quantum mechanics will work. Treat things as entities that have some vaguely wave-like behavior, and figure out when and where they will make standing waves, etc. How should you interpret all that? It doesn't really matter, if the eigenvectors and eigenvalues turn out to be useful then you can use them.
Just use whatever's useful.
