coNP and limitation of NDTM
These complexity classes are not just defined by a machine model but also by the criteria about when a machine accepts a strings. So although $\mathsf{NP}$ and $\mathsf{coNP}$ can be defined using essentially the same underlining machine structure their accepting criteria are different. That is the main point you are missing.

In my experience, generally it is intuitively more clear for people to think about $\mathsf{NP}$ using its verifier-certificate definition and forget about the non-deterministic TM (which can be confusing).

We say a language $L$ is in $\mathsf{NP}$ if there is a polytime DTM with two inputs $V$ (think of $V$ as a certificate verifier) s.t. for all $x$,
$ x \in L$ iff there exists a polynomial size certificate $y$ s.t. $V$ accepts $(x,y)$.

In more intuitive terms, $\mathsf{NP}$ is the set of problems that given an instance and a polysize solution/certificate/proof for that instance, we can efficiently verify the correctness of the given solution.
E.g. given a graph and a list of vertices in it, it is possible in polynomial time to check if the list is a Hamiltonian path in the graph. Therefore Hamiltonian path problem is in $\mathsf{NP}$.

So in intuitive terms, $\mathsf{NP}$ is the set of problems that a given solution can be confirmed efficiently and $\mathsf{coNP}$ is the set of problems that a given solution can be refuted efficiently. For example, a refutation for a $TAUT$ instance is an assignment that makes the formula false.