The large majority of regulators will either be damaged and/or will load the output supply if you connect the outputs directly.
As you say you are using a jumper to power only one regulator at a time you could instead place the jumper in the regulator output circuits. This will mean that all 3 regulators are always powered, but the standby current is usually small compared to load current.
Or you could use an input and output jumper per regulator.
If output jumpers are unacceptable you could place an N-Channel MOSFET in the output of each regulator. Source to load, Drain to regulator output, gate to regulator input. When the regulator is powered the MOSFET is switched on. The MOSFET Vgs_th (turn on voltage) needs to be comfortably LESS than the regulator voltage drop in use. The MOSFET Rdson (on resistance) needs to be low enough to only drop minimal voltage when on at full current. eg a 50 milliohm RdsonFET will drop 50 mV at 1 amp.
Use of Schottky diodes on the outputs as Masterleous suggested WILL work BUT the diodes will typically drop 0.3 to 0.6V and the drop will vary with load. Most Schottky diodes at 1A or so are closer to 0.5V+ than 0.3V so this drop is significant and leads to a variability of outp[ut with load. My series MOSFET concept above does the same thing but the MOSFET acts as a "super diode" with very little voltage drop.
Schottky diode voltage drop:
I frequently see Schottky diode forward voltage drops being quoted as "about 0.3 Volts". This CAN be the case, but it's not usual. To get low forward drop you have to design for it. This usually involves operating the diode close to it's maximum reverse voltage ratings and / or using it at far below its peak rated current. Boty these effects are demonstrated in the table below.
The 1N581x series diodes are 1 amp continuous use Schottkky diodes withy reverse voltage ratings of
1N5817 20 Volts
1N5818 30 Volts
1N5819 40 Volts
At 1A the worst case Vf rating at 25C for each of these is typically 0.45V , 0.55V, 0.60V
Exact voltage quoted varies by manufacturer but few or none quote below 0.45V for the 1N5817. So, if you have an eg 15V circuit, the 1N5817 diode will drop about 0.15V less worst case than the 1N5819. That's about 1% of total voltage and in many cases will not make much difference.
BUT I see these diodes or the equivalent SSxx series used in portable lights and solar powered equipment where they may be used in the output of a boost converter providing eg 3V to drive an LED.
Here 0.15 V is 5% of the voltage used to drive the LED. In a simple DC circuit a 1N5817 driving a 3V LED at 1A would need 3.45V input and 1 1N5819 would need 3.6V (worst case in each instance).
1N5817 efficiency is 3/3.45 =~ 87%.
1N5819 efficiency is 3/3.6 = 83%.
For a given amount on input energy the 1N5817 will operate for (87/83-1)x 100 =~ 5% longer.
Energy losses are 13% and 17% so 1N5819 dissipates (17/13-1)x 100 =~ 30% more energy.
In many cases this degree of difference is not important or even trivial.
In others, an extra 5% runtime can be invaluable. By the time you get to worrying about effects of this order you will also be looking at all other areas in a design that cause losses.
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
V2 will sink very little current unless there's a load attached (similar for V1).
Your two voltage regulators are in parallel, so their branch currents add (see: Kirchhoff's Current Law).
So if V2 is attached to a load which will draw 0.4A at 1.2V, then that means V1 can draw at most 0.4A regardless of what voltage V1 is set to.
Note that linear regulators (like the LM317) have a drop-out voltage, so your output channels will only be able to regulate up to about 2V below the input voltage (the 12V output from the boost converter).
V1 will not be able to draw 10W because the single boost converter is supplying the two regulators; it is only rated to supply 10W total. There is 4.8W drawn by V2 (12V*0.4A, note that this is 12V because V2 is a linear regulator), thus V1 is limited to 5.2W. edit: However, this limit is above the 4.8W limit of what V1 nominally could draw because of the specifications on the output limits of 12V and 0.4A. So even if V2 was not present, V1 could never deliver 10W.