Everything should be equivalent, including the effects on clocks and light.

How about the vertical light clock in the last paragraph of the question. I write it down for you again or you can scroll up.

A light clock installed in the elevator of Einstein consists of two mirrors, M1 (attached to the ceiling) and M2 (attached to the floor) facing each other at a reasonable height "h" apart. Would the bouncing pattern of light beam/ pulse in between M1 and M2 be the same (equivalent) when the elevator is at rest on the ground and moving upward @ the rate of 9.8 m/s/s therein free space away from the local attraction of gravity?

Relative to onboard observer:

Two possibilities:

1- the upward and downward motion of the pulse in b/w m1 and m2 are the same

2- the upward and downward motion of the pulse in b/w m1 and m2 are not the same

which one pattern ( 1 or 2) would you choose in order to make it equivalent to the same vertical clock here on earth at rest and why?

Do we also need a high rise system here on earth in order to notice the delay in upward motion of a pulse and rapid falling of a pulse?

Option 2. That is what should be observed in both cases. But again, it is insignificant.

Note that the math is more complicated in this case, and gets more complex when you throw in relativity. So I will do the times for the outside observer.

If you had a 1 km tall building (reasonable for tall high-rises), then as a first approximation, it would take ~ 3 us for the light to reach the top of the building.

In that time the building would have moved ~0.05 nm, which would take the light an additional 2e-19 s, i.e. basically nothing.

Doing the relativistic correction, you get basically the same number.

For the light path going down, you also get basically the same number, but in the opposite direction (i.e. the light gets there sooner).

However, as the light going up has to go further, more time is spent on the upwards journey and thus the overall effect is a tiny increase in the total time taken.

Technically or practically, once the elevator is stopped then it should stop for both observers but theory says it’s not for one of the observers as either a pulse is moving in the same direction of elevator or opposite to it which will make difference in timing of stopping - Right

Again, not quite.

Practically once the elevator stops, it stops and there is no difference in timing, due to just how insignificant the difference is.

But technically, that is not the case. To start with the elevator can't even stop instantly. Lets assume a simple case of the elevator being thrust upwards by a rocket.

The elevator is then acting somewhat like a spring. The force is compressing it, accelerating it upwards as it does so.

When you remove the thrust, the elevator is still compressed. The bottom stops, but that compression keeps the top accelerating, decompressing as it does so. Eventually the decompression reaches all the way to the top and it has all stopped.

Even if you did manage to have the elevator stop all at once, that would only be the case for a particular reference frame. For other reference frames that will not necessarily be the case.

Let the occupant of the elevator release a ball from his hand. The acceleration of the said elevator increases due to the reduction of weight on its floor.

For an outside observer, the floor of the elevator is catching the ball while the ball stays at its original position. NOTE: the greater the velocity of elevator in its constant acclaration of 9.8 m/s/s the less time it requires to catch the ball

Is the person just releasing the ball, or are they making its velocity 0 relative to an outside observer?

If the former, then the ball is initially travelling at the same velocity as the elevator and it will take the same amount of time to catch up regardless of the velocity of the elevator.

If the latter, then yes it will reach the floor quicker, but that is because it effectively be thrown at the floor.

For an inside observer the ball is falling on the floor of the elevator but would he feel an increase in his acceleration as soon as he releases the ball? Similarly would he feel a declaration when the ball hit the floor due to an increase in weight of elevator again? - here you don't agree with above NOTE but thats ok

This depends upon a multitude of factors, including what is causing the acceleration. If the acceleration is kept constant then there will be no change. However if it is based upon a constant force, then it will increase when the ball is released and drop when the ball hits the floor.

Also note that the same happens with Earth and gravity. When you drop a ball, not only does the ball accelerate towards Earth, the Earth (and you) accelerate up towards the ball, but again, the effect is insignificant.

If the path of a pulse curve down relative to the inside observer then it would eventually miss its target of a mirror. Also, its pattern of bouncing is not similar to the bouncing of pulse on earth as you said. – Again in contradiction with horizontal light clock on ground at rest.

Not in an ideal environment.

In an ideal environment the light initially goes upwards at an angle of a. During its journey it curves downwards eventually reaching an angle of -a.

It then hits a mirror perpendicular to this and goes back heading upwards at an angle of a, curving down eventually reaching an angle of -a, and bouncing off another mirror.

The light will continue bouncing back and forth between these mirrors.

To the outside observer the light would initially be going upwards at an angle of a, but then get steeper and steeper as it bounces between the mirrors.

does the inside observer know that he is moving upward relative to the stationary horizontal beam? no

Only in the sense of it being too insignificant to notice, or it being equivalent to being stationary in a gravitational field.

Also, a thinker simply suggests that an inside observer in the accelerating frame would feel a ground condition if he throws a ball upward and then catches it again after falling. Here a thinker, who made comments on behalf of an inside observer, doesn’t add upward velocity (which he added in order to make a horizontal beam curve down) to the velocity of the elevator and subtract downward velocities of a ball from the upward velocity of elevator which will make a huge contradiction to the ground conditions.

I'm not sure what you mean by ground condition, but if you mean what it feels like to catch the ball, the velocity of the elevator doesn't make a difference. That is because what matters is the relative velocity. That is because during the collision the velocities need to be matched, i.e. a force is applied to accelerate the objects such that both objects travel with the same velocity.

It would feel the same to catch a ball heading towards you at 1 m/s with you stationary as it would to catch a ball heading away from you at 99 m/s, with you travelling at 100 m/s.

Assuming you have a train system where you live with decent stretches of straight track you can test this with a friend by throwing a ball back and forth as the train is moving.

If you stand at rest in a gravitational field and throw a ball upwards at a speed of v, it will take -2*v/g to come back down, and it will be travelling at a velocity of -v when you catch it and thus has a relative velocity of -v.

If instead you throw it in an elevator in free space accelerating upwards at a rate of a, when this elevator is at a velocity of u, then the position of the ball will follow the equation (u+v)*t, while your hand follows the equation u*t+0.5*a*t^2.

The ball and your hand meet again when these 2 are equal, i.e.:

(u+v)*t = u*t+0.5*a*t^2

u+v=u+0.5*a*t

v=0.5*a*t.

i.e. at a time of 2*v/a.

The velocity of the ball will still be u+v.

The velocity of your hand will be u+a*t=u+a*(2*v/a)=u+2*v.

Thus the velocity of the ball hitting your hand, relative to your hand is -v.

This is the same.

The appropriate analogy for a beam of light curving downwards is throwing the ball horizontally, where it will appear to curve downwards, but unlike light it will typically be noticeable.