While everyone agrees that jumping in a falling elevator doesn't help much, I think it is very instructive to do the calculation.
General Remarks
The general nature of the problem is the following: while jumping, the human injects muscle energy into the system. Of course, the human doesn't want to gain even more energy himself, instead he hopes to transfer most of it onto the elevator. Thanks to momentum conservation, his own velocity will be reduced.
I should clarify what is meant by momentum conservation. Denoting the momenta of the human and the elevator with $p_1=m_1 v_1$ and $p_2=m_2 v_2$ respectively, the equations of motion are
$$ \dot p_1 = -m_1 g + f_{12} $$
$$ \dot p_2 = -m_2 g + f_{21} $$
Here, $f_{21}$ is the force that the human exerts on the elevator. By Newton's third law, we have $f_{21} = -f_{12}$, so the total momentum $p=p_1+p_2$ obeys
$$ \frac{d}{dt} (p_1 + p_2) = -(m_1+m_2) g $$
Clearly, this is not a conserved quantity, but the point is that it only depends on the external gravity field, not on the interaction between human and elevator.
Change of Momentum
As a first approximation, we treat the jump as instantaneous. In other words, from one moment to the other, the momenta change by
$$ p_1 \to p_1 + \Delta p_1, \qquad p_2 \to p_2 + \Delta p_2 .$$
Thanks to momentum "conservation", we can write
$$ \Delta p := -\Delta p_1 = \Delta p_2 .$$
(Note that trying to find a force $f_{12}$ that models this instantaneous change will probably give you a headache.)
How much energy did this change of momentum inject into the system?
$$ \Delta E = \frac{(p_1-\Delta p)^2}{2m_1} + \frac{(p_2+\Delta p)^2}{2m_2} - \frac{p_1^2}{2m_1} - \frac{p_2^2}{2m_2} .$$
$$ = \Delta p(\frac{p_2}{m_2} - \frac{p_1}{m_1}) + (\Delta p)^2(\frac1{2m_1}+\frac1{2m_2}) .$$
Now we make use of the fact that before jumping, the velocity of the elevator and the human are equal, $p_1/m_1 = p_2/m_2$. Hence, only the quadratice term remains and we have
$$ (\Delta p)^2 = \frac2{\frac1{m_1}+\frac1{m_2}} \Delta E .$$
Note that the mass of the elevator is important, but since elevators are usually very heavy, $m_1 \ll m_2$, we can approximate this with
$$ (\Delta p)^2 = 2m_1 \Delta E .$$
Energy reduction
How much did we manage to reduce the kinetic energy of the human? After the jump, his/her kinetic energy is
$$ E' = \frac{(p_1-\Delta p)^2}{2m_1} = \frac{p_1^2}{2m_1} - 2\frac{\Delta p\cdot p_1}{2m_1} + \frac{(\Delta p)^2}{2m_1}.$$
Writing $E$ for the previous kinetic energy, we have
$$ E' = E - 2\sqrt{E \Delta E} + \Delta E = (\sqrt E - \sqrt{\Delta E})^2 $$
or
$$ \frac{E'}{E} = (1 - \sqrt{\Delta E / E})^2 .$$
It is very useful to estimate the energy $\Delta E$ generated by the human in terms of the maximum height that he can jump. For a human, that's roughly $h_1 = 1m$. Denoting the total height of the fall with $h$, we obtain
$$ \frac{E'}{E} = (1 - \sqrt{h_1/h})^2 .$$
Thus, if a human is athletic enough to jump $1m$ in normal circumstances, then he might hope to reduce the impact energy of a fall from $16m$ to a fraction of
$$ \frac{E'}{E} = (1 - \sqrt{1/16})^2 \approx 56 \% .$$
Not bad.
Then again, jumping while being weightless in a falling elevator is likely very difficult...
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.
Best Answer
The bus experiences considerable drag, and will therefore fall more slowly than a person inside the bus. The scenario is possible in principle - but after carefully viewing the clip and doing some calculations, I believe that the details are inaccurate.
Assume the bus has a mass of 5000 kg (pretty light for a bus), and is 3 m wide by 3 m tall - so the forward facing area is 9 m2 (it will be more if the bus falls at an angle, but in the movie it seems to be dropping straight. Despite the initial angular momentum as it tips over!).
The drag force is
$$F = \frac12 \rho v^2 A C_D$$
For the dimensions given, after one second the velocity is 5 m/s and this force will be approximately
$$F(1) \approx 0.5\cdot 1.2 \cdot 5^2 \cdot 9 \cdot 1.15 = 135 N$$
(assuming drag coefficient of 1 - close to, but a little bit less than, the coefficient for a cube; That is not yet enough to make the bus go visibly slower.
We need to know how high the bridge is. It turns out that this scene was filmed at Lion's Gate Bridge in Vancouver. This has a clearance height of 61 m. That's roughly what I would have estimated based on this picture (screen shot from the YouTube clip at 1:11, with blocks added by me to show it's about 6 buses high. A typical bus is about 10 m long, so it all makes sense):
Now the actual drop takes from 1:10 to 1:20 in the clip - that would suggest there is some "temporal dilation" going on. Normally, dropping 60 m would take 3.50 s; but in the film it takes 10. This is a clue that normal laws of physics have been suspended for the scene.
For an object free-falling in the presence of drag, the terminal velocity is given by
$$v_t = \sqrt{\frac{2mg}{C_D\rho A}}$$
(About 95 m/s for this bus) and the "characteristic time" $\tau$ (used for the equation of motion)
$$\tau = \frac{v_t}{g}$$
The velocity as a function of height is
$$v = v_t \sqrt{1-e^{-2gh/v_t^2}}$$
This means we can calculate the velocity of the bus and the passenger as a function of height/time: plotting their relative velocity and the position of the passenger relative to the bus gives:
This tells me that the scene as shown in the movie does not follow usual Newtonian physics. Either the air was ridiculously dense, the bus was much lighter than it looked, or... they just did what they wanted because that's what the script called for. Movie physics.
Myth. Busted.