How do simple bird waterers work? I made a simple bird waterer made from two 1.25L PET bottles.
Let's call these two bottles the reservoir and the tray.

Setup
The tray is a bottle with the top cut off to leave only the bottom third. The location of the horizontal slice really only matters to help stabilise the reservoir which will be inverted and placed into the tray.
A window is cut into the tray to allow a bird to duck its head in and take a drink.
The reservoir is a whole bottle (compressed in the depicted image for formatting) with the lid removed and a small hole drilled a few cm's up (when inverted) from the neck. The location of this hole is important as it is a negative feedback loop to control the water level in the tray.
Operation
Fill the reservoir with water. Invert it and sit it in the tray. The water will drain from the reservoir through the neck until the tray is full of water and no more. As water is taken from the tray, more water drains into the tank to fill it. Once the tray is full, the water stops draining and remains held in the reservoir.
See a video of it here:
https://www.youtube.com/watch?v=Qn-1wEuHG7g&t=21
Explanation
The way I thought it would work is the air hole allows air to leak into the reservoir which keeps the air pressure the same as the outside. Water flows out through the neck since the air pressures are matched outside and inside the reservoir. 
When the water level reaches the air hole, the hole is sealed and the 'vacuum' in the top of the reservoir increases until the air pressure pushing down on the water sitting in the tray prevents the escape of any more water.
This is not what happens though. The water stops rising before it reaches the air hole. You can see this in the video - I've annotated a blue box where air hole is.
So the question is - how does this really work? 
I'm thinking it is to do with air pressure but also possibly surface tension. If the bored air hole was too large, intuitively the water would just fall out no matter what.
Why though if the air pressure in the reservoir is lower than outside, doesn't the air continue to push through the air hole until it is sealed?
 A: Surface tension is plausible : it implies that, at the level of the hole, the pressure in water will be less than 1 atm. Thus the pressure isolines will look like this:

Quantitatively, curvature $c$ will be of the order $1/(1 $mm$)$ because the hole looks about 1 mm size in your video, which with pure water would lead to a pressure difference drop of the order of 70 Pa across the interface. Then the flat interface with atmospheric pressure needs to be $c\gamma/(\rho g)$ lower than the hole, which is about 7 mm with pure water. Because you probably have some impurities, this will be less in practice. As @pwf suggests, you can use surfactants (soap) to reduce it (but that won't be a good thing for the birds!)
A: No surface tension, just equalising pressures.the pressure outside the tray pushing on water surface into the air hole is equal to 1atm + water height above the top edge of air hole gdensity of water and this equals the pressure inside the bottle pushing water on the air hole to go outside which is the height of water * g * water density + air pressure above the surface inside the the bottle which is definitely less than 1atm by an amount equal to the pressure of water column above the edge of air hole outside .To see it better the longest the bottle the higher the column of water above air hole is.
