It all depends how much you want to model the problem according to realities. Getting exact answers is almost an impossible task. I was working for a while on noise propagation calculation for traffic and industry sources as a business. Things are extremely complicated, you have dozens of formulas to calculate that. In fact very precise EU and state-dependent regulations exist for calculating sound/noise propagation. Eventually, everybody use special software for that and I guess only those who write software are fully aware of all details.
But as a bottom line: absorption due to air itself (which by the way largery depends on humidity and air temperature!) is much smaller than intensity fall due to geometric reasons (I am talking about earthly dimensions). Even when sound propagates through thick vegetation, the additive reduction according to EU-regulated modelling, is few dB per 100 meter. And bottom line, when doing real absorption calculation in air, exponential formula fails. I would gladly provide you exact formulas from calculation regulations, but they are no longer at my disposal.
As it happens, absorption due to presence of soft materials (like for example earth ground) is more important, and reflections are also a much more acute effect. Other effects are well described in the document you cite. So I think it would be fairly right to say for real problems, that intensity reduces $\frac{1}{r^2}$ for point sources (mostly industrial sources) and $\frac{1}{r}$ for linear sources (mostly traffic sources). If you only explain this difference (point and line sources) I think you shall do enough.
Note: The right derivation for point source (without theoretical absorption) is rather simple
$$A(r, t) = A_0\frac{\sin(k r-\omega t)}{r}.$$
$$I_0 = |A(1,t)|^2 = \frac{A_0^2}{2}$$
$$I = |A(r,t)|^2 = \frac{A_0^2}{2 r^2} = \frac{I_0}{r^2}$$
And here is for linear sources
$$A(\rho, t) = A_0\frac{\sin(k \rho-\omega t)}{\sqrt{\rho}}.$$
$$I_0 = |A(1,t)|^2 = \frac{A_0^2}{2}$$
$$I = |A(\rho,t)|^2 = \frac{A_0^2}{2 \rho} = \frac{I_0}{\rho}$$
There is even much simpler explanation for the formulas above. Intensity is power over area
$$I = \frac{P}{A}.$$
In case if there is no absorption, power remains constant, while area changes. In case of the point source, area increases as $A \propto r^2$, while in case of the linear source, area increases as $A \propto \rho$!
PS: In the document you cite it says "Under 'normal' circumstances, atmospheric absorption can be neglected except where long distances or very high frequencies are involved." This is in full agreement with general experience, as this absorption presents a serious effect in case you calculate distrubtion of noise from huge industrial transformers. Long distance effect is on the other hand completely diminished by other effects.
Is the above reasoning correct to explain why speed of light is constant? (i.e., thinking of it as constancy of any wave in general)
I would say no, although I'm sure you could argue for it if you defined you terms carefully. I just don't think it's good to put light waves and other waves on the same footing. First, light does not need a medium to travel through. Second, according to SR, light behaves much different than waves such as sound in air. I will explain.
Let's say I'm moving near but less than the speed of sound in air relative to the air itself. I emit a sound wave. Since the speed of sound is constant relative to the medium, I will practically be riding right next to this sound wave. In my own frame, I will observe the speed of my sound wave to be very slow, since I am moving just a little less faster than it.
Now let's say I take off in my spaceship near the speed of light. Then I turn on my ship's headlights. According to my own reference frame, I won't be practically riding along with the light I just emitted. I will see it move away from me at the speed of light relative to me.
And this is where SR breaks away from the behavior of sound waves. Sound waves move relative to the air they move through. Their speed is defined relative to the medium. You could argue that the medium serves as an absolute frame. But light does not have this property. There is no absolute frame we can use.
I want to know whether the fact that speed of light has to be constant is a consequence of other fundamental laws of Universe, or is it just a fundamental law itself.
This is held to be a postulate of SR, but whether or not it is a fundamental law is somewhat subjective, and the answer could change based on what we end up discovering about the universe in the future. According to SR it is a fundamental property of the universe. But maybe we will discover more that explains why this happens. Then that explanation will be the fundamental explanation of the universe.
I will note that your title question isn't answerable with physics. Nothing has to be any way. Why does mass have to warp space-time? Why does the universe have to have more than two fundamental forces? There isn't anything saying it has to be this way. But it is.
Theoretically, could there be other waves that could travel through vaccum at a different speed?
I believe the answer here is no, but I have to admit I can't think of a good reason. Even if there was, I'm not sure it would ruin SR. I believe it would just mean that the particle associated with this wave would need to have mass, but then I'm not sure it could propagate without a medium. I'm not sure about all of this though.
Best Answer
It's because the amount of area "covered" increases as the square of the distance. Imagine a sphere, centered on the source, at a radius equal to your eyeball's location. If the source generates X watts (or whatever unit you like) total, the brightness, i.e. the percent of light which hits your eyeball, is X divided by the ratio of your eyball's area to the area of the sphere. Now move back a few meters. Your eye's area is the same but the sphere's area has increased dramatically. Thus the amount of light, or watts, hitting your eye has gone down.
Absorption makes things worse, of course, but is not necessary for the light your eye sees to decrease. There are plenty of exceptions to the inverse-square rule, e.g. collimated laser source, but this should cover your question.