*Escalator= Moving stairs.
Urm, consider your relative velocity wrt to the moving escalator* when you are standing on it still. Zero. When you move, you exert about as much force as you would on a "normal" staircase of identical charecteristics.Therefore to answer you pointwise:
1. Gravitational force decrease with distance.
Consider g aka acceleration due to gravity. It depends upon mass(kg) of Earth(M) and the radius(R)(unit is meter) of the planet and the formula is g=((MG)/R^(2))=~9.8m/s^(2) roughly. G is the universal constant of Gravity and its value is 6.67*10^(-11) Nm^(2)kg^(-2)
Therefore the gravitational force(F) has unit Newton and is calculated by the formula F=mg where m(small) is the mass(kg) of the object(you) and g the acceleration due to gravity.
Note that the g is independent of the mass of the object. Of course, the object(you) also exert gravitational force on the Earth but considering the difference in the scale it's negligible and can be safely ignored. Or don't ignore it. Whatever floats your boat.
Work is defined as Force*Distance and the units is Joule so the actual work you'll be doing in this case would be (mg)(h2-h1) where hn is your difference in initial and final height, m is your mass and g the acceleration due to gravity. Now....the value of g decreases with increase in R, so actually you'll be doing less work when climbing the same distance on a already high situated escalator rather than a "ground level" staricase. But all this is codswallop because the difference in R is negligible, nor is the centre of mass properly defined.
TO the point, there is this thing called power[unit Watt]. Power=Work/Time. Therefore, the power exerted by you would be more if you are trying to cover the same distance(vertical component, obviously) in shorter period of time. May be you try to climb faster while on escalators? I know I do. All things equal, escalators will help conserve your fat more than the regular stairs. If only because you'll be covering the same distance in shorter time. So take the "static" stairs and stay healthy.
2. Double the force of Gravity? And that too while going straight up?! Gravitational Force between two mass bodies is calculated with F= ((GMm)/R^(2)). And since you are already going up and increasing the distance, and since you cannot possibly influence the Mass(M) of the Earth, and G is G, what we can do is increase your mass dramatically, all the while ensuring that as you go up more mass is addded! Maybe we could teleport lots of Ice-cream to you...
Haha, I am so funny.
In case you were trying to refer to acceleration due to gravity aka g as experienced by your body, while travelling vertically, all you would have to do is build an apparatus like a rocket/plane that can provide you with 2g acceleration vertical. The initial speed would be zero, obviously, and after that it would keep increasing wrt time. So either with known rate of acceleration we can calculate speed at t seconds after launch/take-off, or if you want x speed at y distance than we can calculate the rate of acceleration. This is dealing purely with vertical component ie 2d.
3. You haven't specified in what axis you want to feel this 2g force. If it's the net of the horizontal and vertical in 3d, then it can be easily calculated through trignometric formula though because of my lame HTML skills I cannot elaborate more here. However, you can easily experience this much g's on Earth surface itself 2d. Any car that brakes from 100 kmph to 0 within 1.4 seconds. F1 cars with carbon-fibre brakes regularly hit more g's(momentarily, and decreasing) at the end of long straights followed by slow corners.
4. Relative velocity. And depends on the point of observation. As a point object, and certainly if mass is taken into consideration, all your running stunts are pretty much ignorable and "your" speed will be taken as the speed of train(x kmph). Relatively, however, your body is at rest (inertia) and if you start running(y kmph) towards front you will momentarily be covering more distance per unit time(x+y kmph) compared to other pax(x kmph) and are "faster"(by y kmph) than the train, to someone viewing this whole scene from a staionary platform from the outside. That is, until you run out of room and head back to your seat. So you'll arrive at the same time as the rest of the pax.
Gravity, as you know by now, applies to mass. For all theorotical physics problems, we take a point mass(hypothetical), where all the mass of an object is centered, and then calculate. So, your mass is irrelevant, what matters is the total mass and how it's distributed to make sense from an engineering point of view. Because you are travelling at the same speed as your vehicle, your and your vehicle's mass is combined in calculations. But as soon as you jump out, you'll be taken as a seperate point mass and studied as you form a nice trajectory and come to rest. Strictly, the force of gravity is affected only by the variables mentioned in te formula. Draw your conclusions.
EDIT: Improved Formatting. Relatively speaking.