It doesn't define where the "action"/applied force is coming from
It, the diagram, doesn't have to. The source of the force isn't relevant to kinematics. How that force is generated, the internal processes of the force 'applier', the nature of the force itself are irrelevant to the act of balancing all forces acting on an object in free body diagrams (FBD). Only the value and direction of the force is needed.
But I think what you were trying to ask is more subtle and something else - the "where the force...is coming from". Do you mean, say in the first illustration, if the force is coming through the person's arms or whether the person's action force on the wall extends to some finite extent off of the wall perpendicularly, like a literally manifested red arrow?
If so, it isn't. From the perspective of a FBD, the action force is a vector acting $at$ the $point$ of contact. It has no spatial extent when we analyse it. Similarly, the normal reaction force of the wall on the person acts at exactly that point. These forces in fact exist only at that point when we analyse them.
As per Newton's third law they are vectorially equal and opposite. The FBD is
<---------.--------->
^ ^ ^
| | |
Reaction | Action
force | force
|
Pt. where force acts
(contact point)
fig 1 FBD of the point of contact between the wall and the person. To find the FBD of the person remove the action arrow and replace the dot with the person. For the FBD of the wall, remove the left arrow and replace the dot with the wall.
Are forces measured through something or as an action from one object to the next?
As explained above, no. We analyse forces at points where they act. When forces have spatial extent, they become force fields, but that may be an advanced idea for you at this stage.$^1$
I know that if I exert a force, then that force should still be going through my arm.
The force isn't going through your arm in so much as its being generated by the muscles of you arm.$^2$ Either way what pathway this so called "traveling force" takes through you arm to reach the point of application - the contact point with the wall - is irrelevant. What matters is that at that point, there is a force being applied. From the perspective of analysing force balance at that point, its extent and origin don't matter -it only exists at that point.
A question that whether this force applied should be in the free body diagram at all.
Of course, the FBD of the point of contact must include this force, as its being applied to it. Whether it was travelling down you arm or being generated by you r fingertips is not a criteria for exclusion/inclusion from the FBD. All that matters is, "Is this force acting at this point?"
then the contact action/reaction pairs forces are not equal to force applied because the applied force external to the object?
Not true. For e.g, in the case shown, if the left block be labeled $1$ and the right one labeled $2$, and the external $10N$ force be called $F_e$, with coeff. of static friction $\mu$, $F_{ij}$ represent the force on object $i$ by $j$, and the acceleration of the system be $a$ then for mass $1$,
$$
F_e-F_{12}-\mu m_1g=m_1 a ^{[3]} \tag{1}
$$
If the external force isn't enough to overcome friction, and the masses stay at rest then $a=0$ and eqn. $1$ can be rearranged to show
$F_{12}=F_e-\mu m_1g=-F_{21}$
where the last equality comes from the third law. Since both friction and the external force are, well, external to mass $1$, action-reaction forces can in fact be equal to external forces.
$^1$ and the there is the concept of pressure but I recommend you clear this doubt first before studying that.
$^2$ In fact its the palm, no the arm, no the shoulder, no the torso, no the pelvis, no the legs, no the feet, no - the ground, yeah the ground you stand on that helps you push the wall. If the ground wasn't there you would only be able to push off of a wall (and in so doing push it)
$^3$ for mass $2$ its $F_{21}-\mu m_2g=m_2 a$
