Any transformer connected to 240V AC has to have sufficient inductance in the primary so that it isn't taking a large magnetization current - just think of the primary and ignore the secondary for now - imagine you are only making an inductor to connect to the AC - you don't want it taking ten amps just by itself.
Of course there is a technical reason for not taking ten amps - this will almost certainly saturate the core and fry.
So, armed with the details you have on the toroid such as the \$A_L\$ value, this will help you understand how many turns are needed to obtain an inductance of (say) 10 henries. 10 henries will have an impedance of about 3000 ohms at 50Hz and will take a current of about 80mA when connected to the 240V AC.
Then you need to decide if this will saturate the core and fry. You Used \$A_L\$ and a target inductance of 10 henries to tell you how many turns you need to wind then, you can calculate the magneto-motive force (ampere-turns). Then, divide MMF by the net length around your toriod to get H (magnetic field strength, ampere-turns per metre) and you are nearly there.
Referring to the BH curve in the toroid's data sheet and using the value of H just calculated determine what magnetic flux density (B) you will be getting from the graphs normally supplied - if it's more than about 0.4 Teslas then you will likely run into saturation problems.
I'm not doing the math but it's going to be a close run thing as to whether this toroid is big enough to tolerate mains voltage across the primary - ferrite toroids are not normally used as regular AC transformers - they are better suited to a much higher frequency (as per what you get in an off-line switcher) because of this issue.
If you don't know what the ferrite is made of forget it (and I mean that).
Winding the secondary is a cinch in comparison but before any advise is given, technical detail of the toroid material is needed.
It should work. Spreading the heat dissipation between two regulators may make things easier.
You may wish to put an inverse-parallel diode across each of the 9V outputs that you are going to series. The reason is that if you consider what happens when the output is short-circuited - there will be two regulators each trying to drive the other one negative and one will win. That may harm the regulator that's driven minus and it also could harm the output capacitors (if they're polarized types). Something like a 1N5403 would work fine.
Best Answer
No, you cannot "just" anything with a transformer, until you are sure about the make-up of the secondary windings.
For example a 18-0-18 make-up, which is common in transformers for lower power amplifiers, means it has 2 windings of 18V, but they are already put in series inside the transformer. Yet again, if you have {18V} {18V} as separate windings, it is not uncommon to have two wires of the same colour for each winding. If you want to put them in parallel you will need to make sure you connect the right two wire to each other. If you connect them in reverse your transformer will see a very strong short-circuit and 1. consume a lot of power with no output, 2. eventually burn out.
Also be aware that a transformer has no "plus" or "minus". What comes out of the transformer is an AC voltage, meaning it goes from + on one wire and - on the other, to the other way around at 50 to 60 times per second. LEDs don't like reverse voltage at all. So connecting an AC to a LED will always degrade its lifespan, even if the AC voltage is under its breakdown limit.
So you need to rectify the AC voltage. Some care may have to be taken by not taking the full 0.83A as well, but with design margins on the transformer, that's probably less important.
If you just rectify your LED will still blink. If you want to minimize that, the schematic below is the safest bet. If you know what's going on with the transformer you can save on one rectifier by combining the windings before rectifying, assuming they are exactly the same otherwise. If you are not sure, just use the two rectifiers to be absolutely sure you don't make mistakes. The chosen capacitor gives reduced effect on the blinking, but does increase the effective DC voltage. In this case by about a factor 1.33. If you double the capacitor the blinking will be gone entirely and the effective DC voltage will be 1.4 times the AC voltage.
simulate this circuit – Schematic created using CircuitLab
So as drawn: Vdc = 18V * 1.33 = 23.9V (with load!, without load it's the same as the next one!)
With double capacitance: Vdc = 18V * 1.4 = 25.2V