Why don't humans burn up while parachuting, whereas rockets do on reentry?

Re-entry velocity from LEO is $~7,800 \frac m s$, from lunar space it is as high as $~11,000 \frac m s$ [1].

Different books give the terminal velocity of a skydiver as about $56 \frac m s$ or $75 \frac m s$ [2, 3]. The exact value isn't material, but the fact that it is two powers of ten smaller then re-entry velocity, is.

The difference between skydiving and re-entry is that in order to orbit, you need to go very fast sideways. You essentially fall so fast sideways that you miss the ground when falling towards it (see related xkcd). A skydiver, whether they dive from a plane or balloon, has only marginal horizontal velocity and almost zero vertical velocity to start with. The skydiver then accelerates to terminal velocity, which is quite slow compared to re-entry velocity (see above).

In comparison, the Apollo capsule had a terminal velocity of $150 \frac m s$ at $7,300 \ m$ altitude. It is from that moment on that Apollo behaves like a skydiver. Drogue chutes are pulled that slow down the craft to $80 \frac m s$, and then finally the main chutes that slow down the craft to $8.5\frac m s$ [4].

But that is only the very last phase of the flight. You need to somehow slow down from $7,800 \frac m s$ to $150 \frac m s$ first and descent from space deep into the atmosphere. Creating chutes that can both withstand that and are big enough to slow the craft down enough that high up in the atmosphere is simply not feasible from an engineering point of view, and even if it were, it would probably prohibitive from a weight/delta-v point of view.

The Falcon 9 first stage does not have problems with re-entry heating, although it also reaches space. But that stage does not achieve orbital velocity. It only goes about $2,000\frac m s$ at separation, which is slow enough that heating is not an issue when it comes down (see this question on Space StackExchange).

1: Atmospheric Entry. Wikipedia, the free encyclopedia.
2: Tipler, Paul A. College Physics. New York: Worth, 1987: 105.
3: Bueche, Fredrick. Principles of Physics. New York: McGraw Hill, 1977: 64.
4: W. David Woods. How Apollo Flew to the Moon. Springer, 2008: 371.

The distances and speeds involved are materially different. On the scale of a parachute dive, the atmospheric density doesn't change much (and is relatively high). A parachutist quickly reaches a terminal velocity where the drag from the air matches the pull of gravity.

In a re-entry, you're approaching in a much less dense atmosphere, and you're going much faster. At these speeds, drag warms you up much faster. Also, you're plowing into the atmosphere, and that means you're increasing the drag. Between these effects, you see substantially more heating. A parachutist dropping from orbit would have the same issues with burning up.

There are some interesting things that are done regarding reentry maneuvers. The Chinese had one lunar orbiter which skipped off the atmosphere. The idea was simple. If the orbiter were to re-enter our atmosphere directly, it would receive too much heating. Instead, it was allowed to just enter the rarified upper fringes of our atmosphere, bleed off some of its velocity (into heat) before skipping off the atmosphere similar to a stone on a pond. This gave it time to get rid of some of that heat before a second re-entry brought it down safely.

A human parachuting from $h=4000\,\mathrm m$ (cf. http://adventure.howstuffworks.com/skydiving1.htm) needs to get rid of their potential energy $mgh$. If we assume that all this energy is used to evenly heat up the skydiver, who essentially consists of water with its well-known specific heat capacity $c_{H_2O}=4182\frac{\mathrm J}{\mathrm{kg}\cdot \mathrm{K}}$, the temperature rises by $$ \frac{gh}{c_{H_2O}}= \frac{9.81\cdot 4000}{4182}\,\mathrm K<10\,\mathrm K,$$ not enough to incinerate. One might object that the heating would occur mainly on the front instead of evenly, but the skydive is long enough for much heat to be transported across the body or even convected away form it by the surrounding air.

Note that the mass of the skydiver did not enter into the above simple estimate, only the staring altitude and the specific heat.

A re-entering spacecraft, on the other hand, not only starts form a higher altitude (more than 100 kilometers), but additionally has to get rid of its substantial kinetic energy. If we don't want to look up the numbers for orbital speed, let's try from memory. One always hears that one orbit takes about 90 minutes, hence the velocity must be at least (using the slightly smaller Earth circumference) $v\ge \frac{40000\,\mathrm{km}}{5400\,\mathrm s}\approx 7400\,\frac{\mathrm m}{\mathrm s} $, so the energy per kilogramm of mass is $\frac12v^2\approx 55\,\frac{\mathrm{MJ}}{\mathrm{kg}}$. We see that this is a lot, lot more than the mere $40\,\frac{\mathrm{kJ}}{\mathrm{kg}}$ of our skydiver.

The spaceship might get rid of much of the kinetic energy by using its rocket engine, but that would be unwise: It took a whole lot of fuel to bring the rocket into orbit in the first place (or rather: into orbital speed, after all the potential energy is small compared to the kinetic energy in low earth orbit); hence it takes (almost) as much fuel to achieve the same $\Delta v$ upon re-entry. But in order to have so much fuel available in orbit, it must have been transported there in the first place. This is infeasible because of the fuel-to-payload ratio.