Switch high frequency AC on/off at zero crossing

This type of opto-triac is mostly used in mains voltage applications. Due to the limited current capabilities it's often used as a driver for a triac which is the actual switching device. Your requirements are modest, so you won't need that, and you can use the opto-triac to switch your load directly. The opto-triac is a cheaper solution than an electromechanical relay then, so at first sight looks like a better choice.
An important difference between electronic and electromechanical switches, however, is that the latter have a very low on-resistance, while the former always will have a voltage drop when switched on. That's the on-state voltage mentioned in the datasheet. This can be up to 3V, which in a 230V application won't matter much, but if your supply voltage is only 24V AC that's more than 10%. Your load will probably work at 21V, but you'll have to check it.

Repetitive peak off-state current is the leakage current when the triac is switched off. 2\$\mu\$A is a safe value.

Holding current is the minimum load current the triac needs to remain on when the gate is no longer driven. For an average triac your 20mA may be a bit low, but again the opto-triac's 3.5mA is a safe value. (Besides, the gate will be continuously driven, so it's a moot point. It is important in four-component dimmers, where the diac gives a pulse to switch on the triac, after which the triac is on its own.)

Then there is the minimum trigger current. That's the minimum current you have to supply to the LED to switch the triac on, and we'll have to calculate the series resistor accordingly.
Where did you get that 38\$\Omega\$ resistor value? You need figures 3 and 4 to calculate the value for the LED resistor. Figure 4 shows that 10mA is a safe value, and figure 3 shows that at 10mA the LED voltage will be maximum 1.3V. So \$R=\frac{3.3V - 1.3V}{10mA}=200\Omega\$ maximum. Your 38\$\Omega\$ would result in more than 50mA, which is not only more than Absolute Maximum Ratings (page 4), but also more than your microcontroller will be able to supply. So don't exaggerate, and pick a 180 \$\Omega\$ resistor. At lower resistances the current may become too much for your microcontroller's output. If you want more current through the LED (no more than 20mA, never use the Absolute Maximum Ratings!) you may want to use a transistor. Since you'd need a lot of them, consider a driver IC like an ULN2803.

In conclusion I think this opto-triac is a good choice. Alternatively, you may have a look at the MOCxxx series, for instance the MOC3012 needs only half of the LED current, which your microcontroller would appreciate. It doesn't give a nominal value for triac current directly, but from maximum power dissipation (300mW) we can derive that this should be 100mA. (It says peak repetitive surge current is 1A, 120pps, 1ms pulse width.)

An old question, but it's a cool topic for beginners to wrap their minds around, so I'll answer it. To answer the last question first, remember that voltage appears across a load, while current is measured through the load.

It may be easier to visualize the phase lag concept if you think of a capacitor rather than an inductor. You're probably familiar with the fact that when you charge a large capacitor, it looks like a short circuit at first. At the instant of connection, current is flowing through the cap, but no voltage appears across it because, hey, it's a short circuit, right? As the cap charges up, the voltage across it rises and the current through it falls. This is all that's meant when people say that "the current leads the voltage" in a capacitor.

With an inductor, we say the voltage leads the current because at the instant of connection the inductor looks like an open circuit. A perfect inductor connected to a voltage source at time=0 will have the whole supply voltage across it, with no current flowing through it. During the 'charging' process the inductor stores energy in its surrounding magnetic field, which cannot happen instantaneously any more than a capacitor can be charged instantaneously. So the voltage "leads" the current in this case.

What's interesting about an inductor is what happens when the source is disconnected. A capacitor will just sit there at the same voltage, slowly losing its charge over a long period of time if there is no load across it. But with an inductor, the magnetic field collapses as soon as the power supply is removed, and this happens quickly. A recently-disconnected inductor will try to maintain the flow of current through the circuit rather than the voltage across itself.... but wait, there is no circuit anymore, because we just opened it.

A perfect inductor would generate an infinite voltage in an attempt to keep the current flowing. Even an imperfect one can turn a few volts into several hundred for a short period of time after disconnection. This is why a zero-crossing switch is not the same thing as a snubber. The snubber's job is to give the inductor a load it can drive when the source is removed altogether -- usually a capacitive one since you don't want it drawing current the rest of the time. It keeps the voltage from rising to levels that could hose semiconductors, burn relay contacts with arcing, or otherwise cause trouble.