The circuit is wrong.
A 7805 can be used in an adjustable mode as per the diagram you provided - using R1 and R2, BUT it is not a good regulator to use this way as it draws substantial and variable current via its adj (ground usually) pin which leads to poor regulation. Using an LM317 in this manner is much better - it is designed to be used like this.
The DC voltage will be ~= the peak voltage = 1.414 x VRMS_AC.
(1.414 = sqrt(2))
So 12 VAC x 1.414 =~ 17V
17 - 2*x 0.7 = 15.3 volts.
The 7805 or LM317 can have 2+ volts internal drop leaving about 12.5 to 13 volts possible at the output.
Use an LM317 if possible - datasheet here.
Fig 5 in the datasheet is the same effectively as above but the reference voltage is 1.25V.
To protect against the back polarity problem that you mentioned, connect a diode from output to input (usually non conducting). If Vo > Vin then the diode will conduct and protect the regulator.
You have asked a bunch of questions there which all have straightforward answers, but it's a bit much to try to cover them all in detail this space, but let me give some suggestions.
LED light for plant?
First, before proceeding, are you sure that LED light, which usually has a very narrow spectrum (or a few narrow lines), will be suited to plant light? I don't know about this, but it would be worth verifying before going to effort.
How to power and control LEDs
Next, you need a few clues about how to power and control LEDs.
You don't mention what the role of the Arduino will be -- will it be to turn the LEDs on and off, or do you want it to produce gradations of light intensity?
a) If on/off, you'll want an arduino shield that provides a relay or power-transistor which can switch an appropriate amount of current, which I'll get to below.
b) If gradations, you'll need a shield that can control the current in increments. Or, a popular alternative is an output controller that pulses the light very rapidly, controlling the overall light by the ratio of on to off time. This is referred to as "Pulse Width Modulation" or PWM. Again the PWM output switch element (transistor) needs to be rated for at least the amount of current you supply to your LEDs.
Edit: Arduinos usually have some outputs that are referred to as "analog outputs" but are actually PWM, so this capability is built in to the Arduino -- though you would still need to provide an external transistor to handle the current of the LEDs -- see examples online.
Supplying electricity to LEDs.
This is the mildly tricky part. LEDs are specified with a typical voltage and current number. For Cree ML-E: 3.2V at 150mA. So you might think "I'll hook eight of those up to 24 volts, and that'll be about right". Unfortunately, it's not so simple. LEDs have a characteristic whereby if you supply a little less than the nominal voltage, and they pass very little current and produce little light. A little more than the nominal voltage and they pass a great deal of current, and probably burn out.
So you don't want to supply a fixed voltage direct to an LED. Instead, you provide a supply which regulates the current. You'll notice that the LED supply you linked to is described as a constant current source. But you don't need to be that fancy. Instead, you can use a supply with a voltage higher than that needed by the LEDs, and put a resistor in series. Example:
Supply: 5V
LED: requires 3.2V, 0.15A
Voltage difference: 1.8V
Resistor: I = V/R So R = V/I, = 1.8/0.15 = 12 ohms. (And FWIW, P = I * V = 0.15 * 1.8 = 0.27 W, so choose a half watt or better physical size of resistor.)
Yes, you can put a bunch of LEDs in series, so for your example 6 x 3.2 or 7 x 3.2 would be possibilities, and still have some voltage drop left between the LED requirements and the 24 V supply. (You will need to factor in that whatever is switching the LEDs, such as a transistor, will also add some voltage drop to the chain.)
Generally, it is a bad idea to attach LEDs (or chains of LEDs) directly in parallel, because the actual voltage for the nominal current may vary from one LED to another, and from one chain to another. So multiple LED chains should each have their own series resistor.
Power for Arduino
Transforming 24V for use with Arduino: The easy answer here is a 7805 voltage regulator which is super easy to use. There are zillions of references for this on the web, so I'll not elaborate. Couple of things to attend to:
a) 24V -> 5V is a relatively large drop for the 7805, so you will need to attach it to a heat sink.
b) The switching of the LEDs will cause sharp changes in the demands on the supply, so err on the side of using relatively large capacitors with the 7805, and parallel them with smaller caps to help with the high-frequency aspect of the sharp switching. This thread is representative. Capacitor Sizes for 7805 Regulator.
[Edit] I'd neglected to note that the original question asked about Arduino with 7-12V power input, which is because Arduino Uno has a voltage regulator that handles the power from the Power In jack. The Uno can run on 5V from USB (when no power is supplied at the Power In jack), but if you are supplying power to the jack, then as the questioner mentioned, that will need to be 7V or higher. So a reasonable solution would be a 7808 or 7809 to obtain 8 or 9V from 24V.
Best Answer
1) Yes, generally speaking it should be fine to send the 5V to the arduinos and netduino, unless any of them have ADCs or DACs that you intend to use and that are sensitive to voltage source fluctuations. Note that in any case you should include a capacitor at the voltage input pin to each chip. Use wires/traces that are thick enough to transfer the current that you need, and best not to put the chips in series along a single power line (unless the power line is sufficiently large to minimize the voltage drop along it.
2) I've only ever needed to put a single fuse at the input of the entire circuit board. I wouldn't expect any average use-case to need more fuses than that unless you're worried about some IC drawing too much current or feeding the power supply to a daughter board.
3) Correct, as long as you keep the current below what your voltage regulator is rated for. Note that if you are drawing 5A and plan to use the 7805, then you're going to burn (12V - 5V) * (5A) = 35W just in the regulator. You'll need a beefy heat sink. A switching DC-DC converter might be better for this case.