In general you can assume that as long as the ripple is not too bad on a supply then it shouldn't cause an issue for functionality of an IC. In this case the datasheet gives a ripple of 100mV peak-to-peak, BUT you seem to imply that you're taking this voltage and further regulating it down. Passing a signal with a relatively small ripple and putting it through a linear regulator like a 7803 or an LDO will significantly reduce the ripple and cut back on switching noise passed through to the final circuit. So a ripple which is less than 5% of the expected voltage like this, being passed through a second line of regulation, should be fine for 95% of projects, even one using analog components like an ADC. Just make sure to have input and output filtering capacitors on the linear regulator, and not too much (2200μF is the limit for the power supply you gave) (maybe 47uF and 4.7uF (or 220uF and 22uF if you want it really smooth)).
Because the ripple is so small this supply would probably be fine for digital operations if you were using a 5V compatible chip and you weren't doing Analog work, but I digress.
So looking at the datasheet you provided, the operating frequency of the power supply is 132kHz. The chances that the noise from that is going to manage to produce harmonics in the 2.4GHz spectrum is pretty limited, especially at such small power outputs. Where you really have to start worrying about generated EMI is when you have isolating switching supplies, because these will pass power through the magnetic field, and the interface leaks EMI on most supplies I've used. Generally the higher-power the switching power supply the more noise it generates. I've had a 100Watt isolating switching power supply which put so much noise out I had 30V of noise on the frame of my robot, but that's way beyond the scope of this. I think you'll be fine with your usage!
Hope that helps!
I agree with others that switchers are a better choice in terms of efficiency, but they can be somewhat complicated to deal with if you're inexperienced, and there can be lots of weird effects that aren't immediately obvious (precharge sinking, beat frequencies, etc.) that can make life difficult. Assuming you've figured out your power dissipation and know how much current each rail can deliver, if the linears will work for you, stick with them (at least for the first pass).
If you're trying to achieve a variable-amplitude square wave output on your adjustable rail, the chopping may introduce noise into the main 24V rail, which could show up on the other rails. You may want to have an LC filter between the main 24V rail and the regulator input to provide high-frequency isolation, and will probably need extra capacitance on the adjustable regulator output (bulk electrolytic as well as low-impedance ceramic) if you expect the square wave edges to be sharp.
1, 5) There are some dangers with your scheme.
Power dissipation in the linear regulators will be
\$(V_{out} - V_{in}) \cdot I_{out} \$
which is significant, especially for the lower output rails. 78xx-type regulators have built-in thermal protection around 125°C, and (without heatsinking) a junction-to-air thermal resistance of 65°C/W. Your thermal management will be challenging.
Another potential problem - if the series-pass element in any of your low-voltage regulators fails or gets bypassed (shorted), you'll present the full 24V input to the output. This could be catastrophic to low-voltage logic. You should protect your low-voltage rails with SCR crowbars that can sink enough current to put the DC/DC brick into current limit and collapse the 24V rail (they'll need big heatsinks too). Fuses are unlikely to be good protection since the 24V brick likely isn't stiff enough to generate the \$I^2 \cdot t\$ needed to blow a fuse.
2) Whatever floats your boat.
4) Meters aren't huge loads. Just use one of your rails.
3) Correct - all regulators have headroom requirements. If you want the maximum 24V out, you'll need a direct connection, and will have to rely on whatever intrinsic protections the brick will provide you.
Best Answer
The current flowing into your +5V supply is limited by R5. The exact value depends on just how high Vin can go, but even 100V is only going to drive about 10 mA into the supply.
There are two ways to deal with this: either have a regulator that can sink any excessive current (rare), or make sure that the minimum load on the +5V bus is greater than 10 mA. If the existing circuit doesn't always draw at least this amount of current, you can add a resistor as a dummy load — 470 Ω would draw about 10 mA @ 5V.