Why not decide coNP language L with same NDTM used for $L^c$?
I assume that you want to "decide if $x\in L^c$ and output the complement". Tell me if I misunderstand.

$L^c$ is in NP if there's an NDTM accepts $x$ in "at least one" branch iff $x\in L^c$. So $x\in L$, i.e. $x\notin L^c$ means the NDTM rejects $x$ in "all" branches. Think more carefully about these quantifiers. 

NDTM can always find the accepted branch if it exists, but this doesn't means the NDTM has checked all branches. So we can't decide if $x\in L$ with the same NDTM in the trivial way you mentioned.

Edit: If you want to prove $L\subseteq NP$, you should construct "one NDTM" such that there exists a polynomially long path to an accepting state iff $x\in L$. You might think flipping the answer is quite simple, but where's the answer? There doesn't even exist a state representing "NO". This is the problem. You can't add another Turing machine which read the output of $M$ and output its complement either since the whole structure is not a valid NDTM.