why do textbooks never mention this?
Because in order to travel at supersonic speeds, human beings must be enclosed in a rigid metal tube of some sort. Also, these metal tubes they ride in at those speeds generally tend to be insulated against noise from the outside.
As for trying to place some sort of microphone outside said metal tube, the propulsion system would risk drowning out any atmospherically transmitted noise (i.e. noise can be transmitted through the body of the structure).
Now, if you run the math, it still doesn't work quite like hearing it backwards (see the comment by eudoxos). Although you would encounter the soundwaves in "reverse" order, the shockwaves around you would disrupt anything around you as to make the notion of noise from them irrelevant.
If the car with the siren runs you over you won't be hearing anything after it "passes". By definition the line of travel of the sound source must it intersect the observer. But if you write down the expression for frequency vs time as a function of distance of closest approach you will see that in the limit where that distance becomes zero, the step gradient becomes infinite.
For a perpendicular distance $d$ between the path of the source and the observer, with $t=0$ at the point of closest approach, and velocity of sound $c$, you can write the frequency as
$$f'=f\left(1-\frac{v(vt)}{c\sqrt{(vt)^2+d^2}}\right)$$
As $d$ goes to zero you can see the fraction flips from +1 to -1, giving you the "instant" change when the sound source overruns (coincides with) the observer.
For the derivation, see the following diagram, and find an expression for the relative velocity $v\sin\theta$ as a function of time, the substitute into the classical Doppler equation.
The key step to note is that the velocity towards the observer, $v\sin\theta$, can be found from
$$v_{relative}=v \sin\theta = v \frac{vt}{\sqrt{(vt)^2) + d^2}}$$
Here is a plot of the above equation for a few different values of $d$, the closest distance. As you can see, when the object is coming "almost" straight at you (10 cm closes approach), the frequency (nominally 440 Hz) shifts "instantly" (limited by time resolution of the calculation):
Rewriting the Doppler equation for $d=0$, you get simply
$$f' = f\left(1-\frac{v \text{ sign}(t)}{c}\right)$$
To understand this intuitively, you need to think about the fact that when the source is heading straight towards the observer, the amount by which the distance decreases per unit time is the same. So the next wave is emitted from "closer up", which is why it arrives " a little bit sooner than expected from the frequency". But at that point you can reset your timer, and the next wave again arrives "a little bit sooner than expected" - in other words, each crest arrives "too soon", but by the same amount. That is why the pitch does not change as the source gets closer in the case of the head-on trajectory. For any other trajectory, the component of velocity of the source towards the observer decreases as you get closer, and the pitch gradually decreases.
Best Answer
There is a list on Wikipedia.
Radar guns use an optical Doppler effect to measure speed. Their acoustic equivalent is used in medicine, where it's called Doppler ultrasound and used to measure blood flow or other sorts of motion in the body.
Animals that use echolocation can use the Doppler shift to gain information about the motion of their surroundings.
A sonic boom occurs when the Doppler shift shifts a frequency to infinity.
I guess one can continue to concoct scenarios. I wonder whether, when you drop a cat off a cliff, you can hear the pitch of its screaming drop as it accelerates. (It's not all that cruel - cats can usually survive a fall at terminal velocity.)
You could use it to determine which way a whale is swimming if you have two boats, both listening to the same whalesong. You could even use the Doppler shift to gain information about the position of the whale because both the whale's position and its velocity contribute to the observed Doppler shift at any given place. (I don't have any information about this actually being done, but it might be an interesting problem to work out the locus of possible whale locations for an observed Doppler shift.)
I was also curious about whether the Doppler effect gives us information on the motion of the crust that moves during an earthquake. I found this reference which suggests it does.