How to explain the force applied on this object? Sorry for posting such a simple problem but I'm having trouble understanding this concept.
In the picture below, we have a force, $ \vec F_{app}$ acting on two objects, A and B.
Mass of block A: $ M_a = 4kg $
Mass of block B: $ M_b = 6kg$
The acceleration of the two blocks can be found using F=ma
$$ 20N = 10kg * a $$
$$ a = 2m/s^2 $$
If we find the force Fba on block B from block A we get:
$$ \vec F_{ba} = 6kg * 2m/s^2 $$
$$ \vec F_{ba} = 12N $$
What I am confused about here is why don't we say that the total force $ \vec F_{app} $ applied onto block A is also applied to block B (20N). Is the force somehow distributed into the blocks?

 A: To help your intuition in this respect, think about an object of mass $2m$ that is subject to a force $2F$. You know of course that this object will experience an acceleration $a=\frac{2F}{2m} = \frac{F}{m}$.
Now if I told you that this was really two objects of mass $m$ that were separated by a small spacer, and that I was pushing on just one of these objects with a force $2F$, you have to reconcile what you "know" to be the right answer with the new reality of the description.
Somehow, each of the two objects should only experience a force $F$ (not $2F$) in order for both of them to accelerate with $a$.
Looking at the free body diagram:

I will call the left hand object A, and the right hand object B. You see that there is a force of $2F$ on the left of A, and a force $F$ from A to B, with an equal and opposite force $F$ from B to A. Now why is that force $F$? We can figure this out by solving for the unknown force (let's call it $f$).
We know that the net force on $A$ will be $2F - f$, and the net force on B is $f$. But in order for the system to stay together, they must accelerate by the same amount $a$. This can only happen if $2F-f = f$, so it follows that $f = F$.
You can do a similar calculation when the objects have different mass; it "looks like" an accelerating object is "absorbing" part of the force on one side, and can only transmit a smaller force on the other side. Another way to look at it is to say that when you are moving in the frame of reference of A (which is accelerating), you have to account for the "virtual force" that appears in the accelerating frame of reference - this virtual force is acting on your object, and means that there is a difference between the force on the left and the force on the right. Intuitively that is attractive, but it can be a dangerous thing to do if you don't understand what is really going on. 
A: It's a very common error to think that a pushing force somehow ought to be "transmitted through" an object to the far side so that same force is applied to whatever the object is in contact with.
If that were the case, then the other object would push back with an equal force and then the first object would have no net force on it and it would not move.  What really happens is that the force on the far side of the object is weaker than the force you push with (in the above case 12N as opposed to 20N, as you found).
But I think there is a way to satisfy your desire for forces to be "distributed into" the blocks.  Imagine block A to be a set of vertical slices, like a (very stiff) loaf of bread. You apply your 20N force to the leftmost slice. The force on the other side of that slice will be a little weaker than 20N, and just enough weaker to provide the net force that will accelerate that slice at 2 meters per second per second.
Continue through the slices of block A.  The force that neighboring slices exert on each other keeps dropping linearly until you get to the place where the two blocks meet, and there the force will be 12N. As you move through similarly imagined slices of block B, the internal forces will linearly drop to zero on the far side.
So I do think your instinct that forces "distribute into" objects is not necessarily a bad one.  But don't think that a force just simply "goes right through" an object to the other side.
A: If you draw a free body diagram on Block A, and include the reaction force that block B exerts on Block A, you will see that, to have an acceleration of 2 m/s^2 for block A, the net force on block A has to be 8 N, and so the reaction force of block B on block A has to be 12 N.  Similarly, if you draw a free body diagram on block B, and include the force that block A exerts on block B, to have an acceleration of 2 m/s^2 for block B, the net force on block B has to be 12 N, and so the reaction force of block A on block B has to be 12 N.  So, either way, the action-reaction pair of blocks A and B has to be 12 N.
