How does an NFA use epsilon transitions?

Every time you are in a state which has a $epsilon$ transition, it means you automatically are in BOTH states, to simplify this to you: If the string is $epsilon$ then your automata ends both in $q_0$ and $q_1$ If your string is ‘0’ it’ll be again in $q_0$ and $q_1$ If your string is ‘1’, it’ll be only in $q_2$, because if you look from the point of $q_0$, you have a ‘1’ transition to $q_2$, but you have also to look at case you’re in $q_1$(if you were in $q_0$ you always were in $q_1$ also) then there is no ‘1’ transition, so this alternative path just “dies”. Just by looking at these cases its easy to see that your automata accepts $epsilon$, $0^*$, and going from $q_0$ to $q_1$, the only way to reach $q_2$ is $0^*11^*1$, so, this resumes your automata to $epsilon$, $0^*$, $0^*11^*1$ Hope this helped you, if you have any further doubts, just ask!