How is the missing digit calculated?

It is a quite simple method of adding the digits and checking their remainder when divided by $9$. This is because $576$ is $9 \times 64$, so the resulting value would also be a multiple of $9$ and the missing digit would generally be $9$ less the remainder. This is because all powers of $10$ leave a remainder of $1$ when divided by $9$ since $10^n = \left(9 + 1\right)^n = 9^n + n \times 9^{n - 1} + \ldots + n \times 9 + 1$, for $n$ being a non-negative integer, with all of the terms being a multiple of $9$ except the last one. Thus, the remainder when a number is divided by $9$ is the same as when the sum of its digits are divided by $9$.

As for the sums in the examples being $36$, this is not always necessarily the case as you noticed yourself.

Regarding the person dropping $0$ or $9$, please see Mark Bennet's answer for how to handle that. I suspect the choice of $576$ and the limitation of the number to multiply it with being $4$ digits likely have to do with ensuring this issue of $0$ or $9$ can be overcome using some other method. This is likely another divisibility test (in particular, for $64$ (as Mark Bennet suggested) or a smaller power of $2$, as this is the other factor of $576$), but I haven't considered this in any detail.

The digit sum of any multiple of $9$ is always a multiple of $9$. If you iterate the process of summing the digits of any positive integer until you get a single digit you get $9$ or the non-zero remainder you would get on division by $9$. Subtract this remainder from $9$ to get the answer.

There is one ambiguity, which is that this method does not distinguish between dropping a $9$ or dropping a zero. You can get round this by asking the audience to drop a non-zero digit. Or assuming they give the digits in correct order, with just one digit missing, there is a test for divisibility by $64$ which could be invoked.