Is the force initially exerted on an object the same it will exert on another? If I exert a force on object A and it collides with object B. Does object A then exert the same force I initially exerted on it if it was at that time at rest?
 A: Let us assume the case when object A and B are both of same mass, say m.
Now, you apply a force $F_{o}$ for a time, say, $t_{o}$. During this time, object A accelerates and finally reaches a velocity, say, $v_{0}$.
We assume the 2 bodies (A and B) to be a system, there are no external forces (friction and such) being applied as we are speaking in ideal terms, hence we can conserve momentum. Thus, upon collision, the body B acquires all the momentum of body A (as both have same mass and momentum, $p=m\cdot v$)
This process of transferring momentum is what we call a collision.
Collisions are of two types, elastic and inelastic. In an elastic collision, the initial kinetic energy of the system(before the collision) is equal to the kinetic energy after the collision. In an inelastic collision the above statement does not hold true and some of that kinetic energy gets converted into other forms of energy (such as heat, deformations in the material, etc.).
We take the case of elastic collision to make the case simpler.
Now, during a collision, the momentum from one ball gets transferred to the other (as discussed above). This process is not instantaneous (although it appears as such to the human eye).

Let us assume it takes a time $t_{1}$.
During this time, all the momentum is transferred. Now, we know that force (F) is equal to the time derivative of momentum, i.e $F_{inst}=\frac{dp}{dt}\ or\ F_{avg}=\frac{Δp}{Δt}$ where $F_{inst}$ denotes instantaneous force and $F_{avg}$ denotes average force.
Now we can begin to answer the force applied by one ball to the other
Using initial assumptions of the force applied by observer as $F_{o}$, time for which force is applied as $t_{o}$ and velocity reached as $v_{o}$, we get that $F_{o}=\frac{Δp}{Δt}=\frac{mv_{o}}{t_{o}}$.
Now, during collision, we have the same momentum change for object B, i.e. by conserving momentum we see that object B reaches $v_{o}$ velocity and object A comes to rest.
Thus the Force applied (say $F_{1}$ by object A on object B will be $F_{1}=\frac{Δp}{Δt}=\frac{mv_{o}}{t_{1}}$, where $t_{1}$ is the time we assumed the collision takes.
Hence, we may conclude that the force applied is not the same, due to there being a difference in the time of both events (observer applying their force and the collision time).
A: You are looking for a conservation law - a quantity that is the same after an interaction as it was before.
Force is not conserved. However, momentum (mass times velocity) is conserved as long as we include the momentum of all of the objects taking part in the interaction. And energy is also conserved as long as we include all the relevant forms of energy in an interaction (but some of these, such as heat, light, sound waves or energy in chemical bonds, may not be obvious).
A: No. Consider Newton's 2nd law - force is connected to acceleration, not speed:
$$\sum F=ma. $$
When the object has a certain speed, you can't know what force was applied to make it reach that speed. That depends on how fast the speed was reached, and that information is lost.
And again when the object hits something else, the force it applies depends on how fast it decelerates, which depends on many other unrelated factors (softness, elasticity, angle, speed difference...).
The two situations are not connected in any way.
A: Obviously you can apply very different forces on objects in order to accelerate or stop them (which is acceleration under a different name), and you can very gently accelerate something and then stop it violently and vice versa, so there is no link between the forces at both ends of their travel.
No direct link, that is. There is an indirect link though, and maybe that's what you were feeling. If you apply a smaller force at either end, assuming the same velocity during the trip, that smaller force must be applied over a longer period of time and over a longer distance. Both are (inversely) proportional. Half the force, double the distance and double the time.
The general wording would be that the integral of the force over the time it was applied as well as the integral of the force over the distance it was be applied must be equal on both ends: These correspond to the conservation laws of momentum and energy. If the forces are constant, for example when falling a short distance in Earth's gravity, the integral for the energy and for the momentum boil down to simple products:
$E = m*s$ for the kinetic energy and $p = m*t$ for the momentum.
