Electronic – Pulldown resistor on MCU output

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.)

You have two modes.

Input
The input is high-impedance, so there won't flow any current, apart from a small leakage current, which we'll ignore for the moment. Let's say you have a 10 kΩ pull-down resistor. Since there doesn't flow any current into or out of the input there won't be any current through the resistor, and then, due to Ohm's Law there won't be any voltage across it either. So if the low end is 0 V, so will be the input. The controller sees it as a low level.

Output
Whether the output is high or low, it's low impedance, like for instance 10 Ω. Low won't be a problem: the pull-down already made the level low, and the low impedance of the output only enforces this.

If the output is high, the internal 10 Ω resistor and the 10 kΩ pull-down form a resistor divider. The output voltage will then be

\$ V_{out} = \dfrac{10 k\Omega}{10 k\Omega + 10\Omega} 5 V = 4.995 V \$

So the pull-down resistor changes the output voltage only very slightly, thanks to the big difference in resistance.