From some of the texts I read, one definition of NP is: "An equivalent definition of NP is the set of decision problems solvable in polynomial time by a non-deterministic Turing machine." and that we have the following, where $n$ is the length of the input:
$$ \text{NP}=\bigcup \text{NTIME}(n^k) $$
This means that one way a problem can be shown to be $\in \text{NP}$ is if we can construct a non-deterministic TM $N$ using a polynomial time verifier $V$ on a certificate $C$, or:
TM $N$: on input $x$ of problem instance:
1. non-deterministically guess a certificate $C$ given $x$
2. if V accepts $C$, accept
But if I use the definition $\text{NP}=\bigcup \text{NTIME}(n^k)$, wouldn't this imply that $\text{co-NP} \subseteq \text{NP}$, since I can construct a TM $N'$ that can recognize co-NP:
TM $N'$: on input $x$ of problem instance:
1. non-deterministically guess a certificate $C$ given $x$
2. if V accepts $C$ for any branch, reject
3. if all branches reject the guessed certificates, accept
In this case, since all branches of $N'$operate in polynomial time, $N'$ should also be able to solve problems for co-NP in non-deterministic polynomial time.
But since it is not yet sure if NP=co-NP, how should I understand the definition $\text{NP}=\bigcup \text{NTIME}(n^k)$?