I wondered, what if an object is moving with a constant speed? Maybe a
car. If it travels with constant speed and hits me, it still hurts. So
it's force must not be 0
Just to be clear, prior to impacting you the car travels with constant speed because the net force acting on the car is zero, not because there are no forces acting on the car. The force that the drive train applies to the drive wheels of the car to propel it forward equals the sum of all the dissipative forces (air drag, rolling resistance, mechanical friction of the moving parts, etc.) acting on the car in the opposite direction, for a net force of zero.
Once the car hits you, it is no longer traveling at constant speed. There will be a change in its velocity, $\Delta v_{M}$ and momentum, $M\Delta v_{M}$, where $M$ is the mass of the car and $\Delta v_{M}$ is the change in its velocity. In order for this to happen you had to exert a force on the car to decelerate it and it exerted an equal and opposite force on you per Newton's third law to accelerate you.
The change in momentum is called impulse which in turn equals the average impact force applied to the car times the duration of the impact, $\Delta t$, or
$$F_{average}\Delta t = M\Delta v_{M}$$
$$F_{average}=M\frac{\Delta v_{M}}{\Delta t}$$
Per Newton's third law, the object applies an equal and opposite impact force on you, resulting in a change in your momentum and accelerating you as well,
$$F_{average}=m\frac{\Delta v_{m}}{\Delta t}$$
where $m$ is your mass and $\Delta v_{m}$ is the change in your velocity.
One more thing related to this problem. If the same car is pulling
another car behind it, if the car goes with constant speed, it implies
that the first car doesn't exert any force to the second car at that
moment, since it is not technically pulling it anymore (espacially if
we don't take into consideration the friction).
If the car being pulled is going at constant speed it simply means that the net force acting on the car being pulled is zero. That does not mean the first car doesn't exert a force on the second car. It simply means that the force exerted by the first car equals the sum of all the dissipative forces acting on the second car in the opposite direction. It is the same reason the velocity of the first car was constant before the collision.
So my question would be the following: How do these two phenomenons
describe the same law of motion?
They describe the same law of motion when Newton's second law is stated in terms of momentum. That is, the net external force equals the change in momentum of a system divided by the time over which the momentum changes, or
$$F_{net}=\frac{\Delta p}{\Delta t}$$
$$p=mv$$
In the case of the car initially traveling at constant velocity, and the one car towing the other with both having constant velocity, the change in momentum is zero (no change in velocity) and so the net force acting on each, and both as a system, is zero.
In the case of the car impacting you, there is a net force acting on each of you (the car on you and you on the car) resulting in a change in momentum of each of you.
PS: If gravity wasn't behaving as some kind of acceleration pulling me
constantly to the Earth's core, then if I jumped upwards, would I
start slowly getting away from Earth?
Yes.
If there were no gravitational force and you jumped upward you would exert a downward force $F$ on the Earth and the Earth would exert an equal and opposite upward force $F$ on you per Newton's third law. The force you exert on the Earth of mass $M$ will give it an acceleration of
$$a_{M}=\frac{F}{M}$$
with respect to the center of mass of the combination of you and the Earth.
And the equal and opposite force the Earth exerts on you of mass $m$ will give you an upward acceleration with respect to the center of mass of
$$a_{m}=\frac{F}{m}$$
So you and the Earth will separate. However, the position of the center of mass of the combination of you and the Earth will remain the same in order to have conservation of momentum of the system.
Hope this helps.
