This arises because of the asymmetry in definition of a non-deterministic Turing machine.
An NDTM accepts a language $L$, if at least one of the computation paths results in a "yes". It could so happen that there are other computation paths which result in a "no". But all we need is a single yes.
To get a "no", all the computation paths must result in a "no".
The Complement language requires at least one "no" for a "no", but all "yes" for a "yes".
So given an NDTM $M$ for $L$, you can construct a TM which runs M, and flips the output, but that is not a non-deterministic Turing machine, by definition.
If you take $M$, and just flip the accepting and rejecting states, then you won't get an NDTM for complement of $L$, as for some strings in $L$, $M$ could result in some paths saying "yes" and some paths saying "no". You will put such $x$ in the complement of $L$.
Thus it is not clear that NP = CO-NP, and it is in fact, an open problem.