Gravity on supermassive black hole's event-horizon

With the proper definition of the meaning of gravitational acceleration, the questioner is correct and the other answers that claim that the gravitational force at the event horizon is infinite are wrong.

Per Wikipedia:

In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which must be treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity. Because of this, a renormalized value is used that corresponds to the Newtonian value in the non-relativistic limit. The value used is generally the local proper acceleration (which diverges at the event horizon) multiplied by the gravitational redshift factor (which goes to zero at the event horizon). For the Schwarzschild case, this value is mathematically well behaved for all non-zero values of r and M.

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Therefore the surface gravity for the Schwarzschild solution with mass $M$ is $\frac{1}{4M}$

So with this definition, the OP is correct that the suitably defined surface gravity at the event horizon decreases as the mass of the black hole increases.

Now this surface gravity does not mean that a rocket engine that can produce that acceleration will enable you to hover at that distance from the black hole. It does take an infinitely powerful rocket engine to hover arbitrarily close to the horizon and, of course, no rocket engine could let you hover inside the event horizon.

However, if both observers, A and B are freely falling in from infinity, nothing at all unusual will happen as first B and then A (one meter later) crosses the event horizon. Neither will lose sight of the other at any time. What actually happens is that the photons bouncing off of B as he crosses the horizon will be frozen at the horizon waiting for A to run into them at the "speed of light". B who is inside can toss the ball to A who is falling in but is currently outside the event horizon and A will catch the ball after he crosses the event horizon. This is true since to first order A and B, when freely falling are in a common inertial reference frame and they can do whatever they could do when far from the black hole.

The problem comes if they try to hover with one person inside and one outside the horizon. That is not possible – the person inside cannot hover at all and the person outside would need a very powerful continuously firing rocket engine to try to hover. But then all the effects of time dilation etc. will be occurring for both of them and all the problems noted by the other answers will be true.

Read these questions and answers for more insight:

• How does an object falling into a plain Schwarschild black hole appear from near the black hole?

• How does the star that has collapsed to form a Schwarschild black hole appear to an observer falling into the black hole?

The horizon, for any static black hole, is the surface where the escape velocity is $c$. Thus, your notion that gravitation is weaker at the horizon for larger black holes is incorrect.

EDIT: Consider that the thrust required to hover goes to infinity at the horizon regardless, i.e., the proper local acceleration for a stationary observer goes to infinity at the horizon.

EDIT 2: In your thought experiment, you write that person B is $1m$ inside the horizon. But, it's crucially important to understand that, within the horizon, the radial coordinate is timelike with the future towards the singularity and the past towards the horizon. Person B can no more throw a ball towards person A outside the horizon than person A can throw a ball into the past.