You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
I know you said you don't want to run outlets everywhere, but I think a pile of wall warts is exactly what you want. Jameco has a large selection. 9 V sounds pretty good and that is one of the standard voltages, but you might want to consider 5 V (see below). Get something in the 500 mA to 1 A range.
These supplies are inherently isolated from the line, are usually short-circuit protected (check to make sure, you definitely want that), and draw so little AC power that you can string a bunch of outlet strips together without harm.
In the end, I think this will cost less and provide a better experience for the kids. I remember when I was a kid tinkering with this stuff how frustrating it was to have batteries run down, especially when you're not aware of when you are asking a lot from them. You also feel a lot less guilty abusing a power supply than running down consumable batteries.
You can start a fire with almost anything. If you do just the right thing, even a 9 V 500 mA supply can catch something on fire, but no more so than a 9 V battery and without the chance of the battery itself doing something bad and causing chemical burns.
If you are worried about LEDs getting damaged by them getting hooked up backwards, maybe you should get 5 V or 6 V supplies. Most LEDs can handle 5 V in reverse, and some 6 V. With 5 V supplies, you can eventually run logic cicuits directly without having to use a linear regulator. Lots of stuff will run well from 5 V. If you get 5 V supplies, get at least 1 A capability. That will be useful for running small motors. 5 W total power really isn't all that much.
Best Answer
That would be an isolating DC-DC converter and would be at least as big as the wall-wart which, presumably, has been sized for the load so you would lose the benefit.
Some local capacitance might be a good idea but, in theory, shouldn't be required.
This depends on the transient response of the power supply and should be specified on the power supply datasheet.
One thing that you are assuming is that they all share a common ground connected to the 12 V common. This may not be the case.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Any device with full-wave rectifier reverse polarity correction will have its internal ground raised by 0.7 V.
Introducing a device with input protection as shown in Figure 1 could result in problems. The internal ground will be 0.7 V above your common 0 V line. Connecting, for example, a USB lead to another device may correct the problem but now all the power return will be through the USB lead. If this isn't rated for the current the voltage may rise on that line. You could spend a lot of time debugging this and potentially causing some damage.
Proceed with wisdom and caution.