Let’s say a person walks at 4km/hr. Let’s say at moving airport walkway moves at 4km/hr. A human walking on that walkway would be moving at — well, reason would say 8km/hr, but that seems way too simple for science/math.
That actually is the correct calculation. As long as air resistance is negligible, and for a person walking at 8 km per hour relative to the surrounding air it is, then adding the numbers up is the right way to calculate it. Of course that’s for a simple system of both of them going in the same direction (or subtracting if they’re going in the opposite direction). If they’re moving at a different angle relative to each other then you have to use some geometry.
Don’t we also have to assume that neither is going anywhere near the speed of light?
And that the thing on the treadmill walkway isn’t an airplane?
Yeah, there are lots of complications that can come up, but for the simple case of an airport walkway, none of them are large enough to be relevant.
Yes. But unless we’re dealing with a vacuum, air resistance and friction between your shoes and the walkway will become an issue long before that 
.
Since the OP asked for a formula, the usual non-relativistic formula for the final velocity v, given two initial velocities v1 and v2 (both in the same direction) is simply
v = v1 + v2
The “real” formula, that works for any velocities, is
v = (v1 + v2) / (1 + (v1v2 / c2))
where c is the speed of light. When both v1 and v2 are much less than c, the denominator is very close to 1, so the simple formula gives a result very close to this one.
I wouldn’t call the relativistic formula the “real” one. All it gives is the illusion of precision, and illusory precision is a bad thing. Long before relativity becomes relevant, many, many other things are going to become much, much more relevant, and so if for some bizarre reason we need that much precision, then we need to take all those other factors into account in the equation as well.
I’m not sure what you mean by that. Of course it may be difficult or impossible for a particular object in a particular environment to actually reach a relativistic velocity. It wouldn’t be appropriate to use it for calculating the speed of slow moving objects, like someone walking on a peoplemover, or cars on the freeway. But IF an object has a velocity close to c, this is the correct formula.
Of course the answer is 8. Here is a more complicated example where you actually do have to figure something out. If you swim at a steady1 MPH a mile across and a mile back on a quiet lake, it will take you 2 hours, obviously. But suppose you do the same in a river whose current is 1/2 MPH. You might think that it would still take 2 hours, but that is wrong. The underlying reason it is wrong is that you spend more time at the lower speed.
Let’s calculate. Suppose you start out going downstream. Now your effective speed is 1.5 MPH and you clearly need 40 minutes to a mile at that speed. You turn around and start swimming upstream. Now your effective speed is just .5 MPH and you need 2 hours to go one mile.
This is the reason that to set a world record for 100 meter dash, the windspeed in the direction you are running must be below a certain limit, but there is no such requirement for the 400 meter race since it is all the way round the track and, just as in the swimming, the total net effect of any wind is to slow you down.
Of course you are assuming that the person is walking in the same direction that the walkway is moving. And, if they are not, are they really walking?
If not, then do vector math on the sidewalk motion vector and the human perambulation vector.
Also, assume a spherical cow human.
Right, and the context of this question is someone walking on a peoplemover. So that equation is wrong for this context.
As long as “walking” is well-defined. If you are completely out of contact with the machinery (as you are for a period of time if you are doing what would more often be called jogging or running) then a moving walkway, which imparts speed by propelling you through frictional contact with your foot, is not adding anything to your velocity. So it doesn’t scale much beyond that 4 km/h pace without becoming more complicated.
Is it “wrong” or just needlessly complicated? It’ll still give you the right answer, and if you’re using 4 as your velocity, that’s just one significant digit, so you’ll end up with the same answer as adding 4+4. No need to use it, but I wouldn’t call it “wrong.”
Unless you’re jumping right over the walkway without actually contacting it at all, it’ll still add the same way when you’re running as when you’re walking.
I would argue that that level of needless complication is wrong. By including terms so small, you’re implying that the whole formula is precise enough that those terms mean something. And that implication of precision is incorrect.
There will be a very minor additional loss of velocity from air resistance when out of touch with the (moving) footing. You probably won’t be able to run as fast on a moving sidewalk, due to the stronger headwind. But otherwise, it’s just straight addition.