You are trying to figure out the force your bicycle can withstand, although you have posed the question so as to get that answer back as a "mass multiplier." The answer to your question really varies a ~LOT~ with technique. The best we can do here is give boundaries to the possible set of answers.
Assuming anything approaching good technique, the bike will land and start a rebound without bearing any appreciable weight from the rider. So mass of the bike ('m2') may be ignored.
Forward velocity 'v' and distance 'd' both have nothing to do with the problem.
What does matter is h, m1, hardness of the ground, and technique (to include rider's physical fitness and anthropomorphic measurments). Technique is the ability to decelerate the downward velocity of m1 to 0, while exerting the lowest peak downward force on the bike. Once you figure out what this peak force is, you have "effective" weight at landing.
Assume hard landing surface, so we can remove that from the equation. Assume a simultaneous two-wheel landing (I am not saying this is best, just makes calculation a bit easier). Assume that the tires plus your body mechanics give you 0.3m (just over 1 foot)of deceleration distance for m1. You did not give a weight, so to make math (and subsequent scaling) easy, I will assume 100 kg. y = 0.3m is height of m1 at touchdown and y=0 is height of m1 when your face and/or crotch smash into the bike frame.
F=ma. Or, more completely in this case: (The sum of forces) = ma. In the space of 0.3 m, m1 must come to rest (vertically). To do that you need to generate an 'a' which will result in dh/dt = 0. This in turn will require a net force upward. We need to find dh/dt at (wheels) touchdown; then calculate a; then derive a net force to accomplish that. Three problems. here we go...
1.) dh/dt at contact was given by dbrane (sign correction add for my frame of reference) --> -SQRT(2gh). So if h = 2m, then dh/dt = about -6.3 m/s at wheel contact.
2.) To bring dh/dt to 0 in the space of 0.3 m, we need to use the same form of the equation used in step one, but this time: -(dh/dt) = SQRT(2a(-dy)). For our 2m jump, this is -(-6.3 m/s) = 6.3 m/s = SQRT(0.6a). 'a' = about 66 m/(s^2), which is equivalent to about 6.7 gravities.
3.) We are back to F=ma, with a twist. Define the amount of force exerted by the rider = Fr. Define the amount of force exerted on the rider by gravity = Fg = (100Kg)(-9.8m/(s^2)) = -980N. Then (Fr + Fg) = (100kg)(66m/(s^2)) = 6600N. Fr = 6600N + 980N = 7580N.
The force exerted (assuming the rider can apply it smoothly) is about 1700 pounds force.
By doing a one wheel landing, you could simultaneously shorten the effective drop and lengthen the subsequent deceleration distance; both of which would significantly lower the required deceleration forces. My advice: Limit your jumps to under 2m vertical drop.
It is difficult to determine which will fare better.
Small mammals can survive a fall from arbitrary distances. Here's one article I found talking about cats. A chief contributor to small mammals' survival is that they have a lower terminal velocity due to the way wind resistance scales. Wind resistance scales with the area of the animal, while weight scales with the volume, so large animals fall faster because they have a higher volume-surface area ration. On a long fall (hundreds of times the animal's body length), body size influences impact velocity
Even without this effect, small animals may have an advantage. In humans, we can study the statistics of plane crash survivors. According to Wikipedia "Since 1970, two-thirds of lone survivors of airline crashes have been children or flight crew." Children make up a small percentage of passengers, so it follows they have a better chance of surviving. It's no great leap to attribute this to their body size. However, a BBC news story has an expert saying there is no physiological advantage to being a child when it comes to surviving a plane crash.
I have occasionally seen this sort of question addressed with dimensional analysis, but the difficulty is that it's difficult to pin down what you want to try to scale. The peak force or pressure, the energy dissipated per unit mass, the peak power dissipation per unit mass?
Here's an example argument:
If we assume the two people are impacting at the same speed, they need to dissipate the same amount of energy per unit mass. Assume that people can dissipate a certain amount of energy per unit mass in a given time without harm. Then whoever can make the impact last for a longer time will fare better. A taller person can bend their legs through a longer distance, and therefore can make the impact take a longer time, and therefore can fare better.
However, the assumptions in this argument would need to be verified before we can take it very seriously.
Here's another argument:
When two people hit the ground at the same speed, the time it takes them to stop is proportional to their linear dimension because this time is roughly their height divided by the speed that mechanical waves move through their body. Their acceleration is inversely proportional to height. Their mass is proportional to the cube of their height, so the force is proportional to the square of their height. That makes the pressure independent of height, so large and small people will fare equally well.
Here's another:
Same as above, but the deceleration time is proportional to the square root of height because they're flexing their knees, and so the stopping distance is proportional to height. This now favors short people.
Another:
Same as above, but mass scales with the square of height, because people are not scale invariant (the BMI uses an exponent of two). This favors tall people.
Another:
Same as above, but mass is independent of height because we're considering a skinny twerp and a muscular jock. This now favors light people.
My conclusion is that the problem is indeterminate. It depends on whether we're talking about a scaled-down version of the same person, or a single guy who starts taking steroids to prepare for a parachute jump. It also depends on various material properties of the human body, and on what sorts of things cause injury. Ultimately I think it's an empirical question, or at least one that requires extensive computer modeling.
Best Answer
The person hits the ground at the same speed in both scenarios.
Once you're in the air, you fall towards the ground with a constant acceleration of about 10 m/s^2. Everything falls the same way - rocks, cannonballs, people - regardless of size or somersaulting.
There may be some small effects from air resistance, but not enough to be noticeable. The sommersaulter may also land with a different orientation so that his center of mass falls a further distance. Then he could hit the ground with slightly higher speed. But basically, everything falls the same way.
Check out this Youtube video of a feather and hammer falling simultaneously on the moon, which shows that heavy and light objects fall the same way absent air.
Also see this Youtube video of the TV show Mythbusters dropped a bullet straight down and firing one from a gun. The bullets fall in the same amount of time, regardless of their horizontal speed.