I've read that gravity doesn't affect an object's horizontal motion or vice versa, and some people told me that both balls will reach the ground at the same time. How is this possible, isn't that when a ball is thrown its resultant force changes (using basic vector, gravity and its horizontal force), thus causing a change in acceleration? But how is it possible that the ball reaches the ground at the same time?
Let's say that you throw the first ball with a horizontal velocity component $v_x$ and vertical component of $0$. The other ball is simply dropped.
What happens if you start running at $v_x$ in the same direction as the thrown ball instantaneously after you release it? It appears to have no horizontal velocity, and just falls downwards, accelerating from a vertical velocity of $0$. Its motion appears identical to that of the dropped ball to a stationary observer: it appears to reach the ground at the same time, though you aren't messing with it in any way.
Assuming air resistance is negligible and the ground is level, they will hit the ground at the same time.
In the Principia, Newton attributes this very experiment to Galileo:
By the first two Laws and the first two Corollaries, Galileo discovered that the descent of bodies observed the duplicate ratio of the time, and that the motion of projectiles was in the curve of a parabola; experience agreeing with both, unless so far as these motions are a little retarded by the resistance of the air.
It is a classic demonstration that when two or more dimensions are involved, the motion in each dimension is completely independent of what happens in other dimensions. For this experiment, the two balls have very different horizontal velocities, but that does not affect the vertical motion. Both balls start with zero vertical velocity, and accelerate at the same rate $g$.
There are ways to elaborate the experiment so they don't hit at the same time. For example, make air resistance significant, have an uneven floor, make it curve around the horizon (see Newton's Cannon), move at relativistic speeds, and so on. These elaborations are interesting considerations, but the point of the original experiment is to demonstrate to a learner the independence of the dimensional axes.
If the ball is thrown exactly horizontally, then it will hit the ground at the same time as the dropped one - but it will a lot further away from the thrower.
Where you are going wrong is in assuming there is a horizontal force. The force of throwing the ball imparts linear (horizontal) momentum to it, which is preserved until the ball hits the ground when it is converted into heat, or transferred to another object.
Once the ball has been thrown (and ignoring air resistance) there is only 1 force acting on the ball: gravity. That is the only force that causes the ball to drop. The hand can no longer impart any force, as the ball has left it.
As gravity is the only force in both cases, both balls will hit the ground at the same time.
When you throw a ball, we are assuming that you are throwing it horizontally. This means you are giving the ball only horizontal velocity. It’s initial vertical velocity is zero, as for the case of simply letting go. Since vertical displacement, velocity, and acceleration are all the same, time has to be the same. However, all of this can only be true if air resistance is negligible, that is, there is no horizontal deceleration.
Consider a ball thrown with a horizontal velocity $u_x$. Let the vertical velocity of the ball after time $t$ be $v_x$. Neglect air resistance.
The horizontal motion of the ball is independent of its vertical motion because the force of gravitation is acting perpendicular to the horizontal direction. The ball will accelerate in the vertical downward direction due to this force but the horizontal component of velocity $u_x$ will remain unchanged because no component of force is acting in the horizontal direction.
Since the vertical component of velocity is the same in both the cases, so the balls will hit the ground at the same time.