A falling object initially speeds up. The drag force, $$D=\frac12 C_dA\rho v^2,$$ thus gradually increases. At some point the increasing drag force becomes equal to the close-to-constant weight, $w=mg$. Then whichever speed the object has reached as this moment doesn't change anymore - it stays constant from here on. This is called the terminal speed. So yes, the NASA quote is correct.
If the drag force quickly becomes equal to the weight, then a lower terminal speed has been achieved because the object accelerated for a shorter time. So, due to varying drag forces, different objects will reach different terminal speeds.
A parachute, for instance, has a large frontal area, $A$, and a shape that effectively "catches" the air, causing a high drag coefficient $C_d$. The drag force is thus higher and becomes equal to the weight of parachute-plus-person at a much lower speed, $v$.
To your example, an incoming comet typically enters Earth's atmosphere at a fairly high speed, $v$. When drag starts to have an effect, the enormous speed causes an almost immediately enormous drag force* (note that the speeds is squared in the drag formula). This is why its front start burning with the enormous fluid friction it experiences. While burning at the front, it simultaneously is slowed down rapidly. Soon, the drag force is low enough to equal its weight, and then constant (terminal) speed has been reached.
These two scenarios - a parachute and a comet; something dropped from rest and something entering at high speed - illustrate the two different approaches to terminal speed: either the object starts at a lower-than-terminal-speed and speeds up until the drag has increased enough to balance out the weight, or the object starts at a higher-than-terminal-speed and slows down until the drag has reduced enough to balance out the weight.
In the former case, you see speeding up until constant speed, in the latter you see slowing down until constant speed. Only in the latter case can you see burning at the surface.
* Note that the air density, $\rho$, is lower (the air is thinner) high up and increases as the objects come closer to the ground. This would increase the drag force gradually, meaning that the drag force would become larger than the weight. The object will thus decelerate a bit and slow down a bit until they again balance out. This happens gradually. In other words: the terminal speed changes as the object falls - this is clear from the terminal speed formula which includes the air density:
$$\small{v_\text{terminal}=\sqrt{\frac{2mg}{\rho A C_d}}.}$$
The density quickly becomes fairly even and constant when closer to the ground, so the low-altitude skydiver might not experience such changing terminal speed but will indeed fall at a practically constant terminal speed.**
** In fact the gravitational acceleration, $g$, changes as well with altitude, which would also change the terminal speed gradually. But out atmosphere is not that thick (compared to the size of the planet and the extend of the change in $g$), so the density change happens much faster and is much larger than the $g$-change through the atmosphere, making the $g$-change often practically irrelevant, although it might be included in some delicate calculations.
