Point of application when the force is the up thrust? We have been taught that the point of application is the point at which the force is applied. In contact forces, the point of application is the point of contact and in forces acting from a distance, the point of application is the center of gravity of the object on which the force is done.
So for example consider a ball hanging from a string. The force on the ball done by the string is the tension force (a contact force) and the point of application is the single point of contact between the string and the ball. So if I want to represent the force, I will have to draw an upward vector where the tail of the vector starts from  that point of application.
Now consider the example of a ball fall under the effect of its weight. The force here is weight which is a force acting from a distance The point of application in this case is the center of gravity of the ball which is mathematically the center of the sphere. So if I want to represent the force weight, I will draw a downward vector where the tail of the vector starts from the center of that sphere.
Now the confusion arises when it comes to archimedes up thrust force. We know the force is a contact force, but in this case the point of application is that of forces acting from a distance, i.e. the center of gravity. 
I am confused, how is that the case?
 A: In the case of buoyancy, the net upward force comes from differential pressures on the outside of the object.  Everywhere the surrounding fluid is in contact with the object, the pressure of the fluid creates a force on the object.
Because we assume the object to be rigid, we can sum all these separate pressures and find that they are equivalent to an upward force acting on a line through some point (not necessarily the center of gravity).  If the object is not sufficiently rigid, this simplification might not work.  The object could instead collapse or deform.
The same thing is true with gravity.  In reality, every piece of the object is being tugged by everything else around it.  But with the assumption of a rigid object in a nearly constant gravitational field, the net force is equivalent to $mg$ acting through the center of mass.
A: The statement that "...in forces acting from a distance, the point of application is the center of gravity of the object" is not technically correct.  (I wish I had a copy of this textbook so I could see if the author explains this over-simplification somewhere else.)  What the author means is that the sum of the individual forces on all particles of the object can be mathematically modeled as if it were a single force acting on the center of gravity. 
To see why the original statement can't be true, consider that some objects have a center of gravity that is outside of the actual object. (For examples, see https://www.quora.com/What-are-examples-of-when-the-center-of-gravity-of-a-body-is-outside-the-body or slide 4 on this page: https://www.slideshare.net/ElviIdiosolo/center-of-gravity-40049394.  If the force literally acted only at the center of gravity, then the force would be acting on empty space and these bodies wouldn't be affected by the force (e.g, gravity) at all.
So, as BowlOfRed already mentioned, the bouyant forces truly acts everywhere the surrounding fluid contacts the object.  But in your diagrams, you can model this as if there were a single force acting at a single point.
Incidentally, even contact forces (on a rigid body) could be modeled as if the vector sum were a single force acting at a single point of the object.  So, in my humble opinion, the distinction between contact forces and forces acting at a distance—at least in the case of rigid bodies and vector diagrams—is really not needed and only adds confusion.
