To answer just the part about 2-digit numbers:
Other answers have explained why the result of subtracting the digits of a number $n$ from $n$ is always divisible by $9$ and that if a number is divisible by $9$, so is the sum of its digits.
If $n$ only has two digits, it can't be bigger than $99$. So when you subtract its digits, you get something that's (i) less than $99$, and (ii) divisible by $9$. The only possibilities are $9$, $18$, $27$, $36$, $45$, $54$, $63$, $72$ and $81$.
Now, why do their digits always sum to 9? Two ways of looking at it:
First way: The biggest sum you can get from adding two digits is $18$, when both digits are $9$. But that needs the result of your first step to be $99$, which we've said it can't be. But we know it's a multiple of $9$, meaning its digits add up to a multiple of $9$. Since they can't add up to $18$, the only available multiple of $9$ is $9$.
Second way: How do you get from, say, $18$ to $27$? To add $9$ you can add $10$, which increases the first digit by $1$, then subtract $1$, which reduces the second digit by $1$. When you add the digits, the two changes cancel out and the digits still sum to $9$. Thinking of $9$ as $09$, this works all the way up from $9$ to $90$ and only goes wrong when you get to $99$—which is too big to be one of the options so the sum is always $9$.