Why isn't NP = coNP?
Solution 1:
That works for deterministic machines, but not for nondeterministic, since a nondeterministic machine has many "branches" in its calculation, and a language is defined to be accepted by a machine if there exists a branch that accepts it (in the normal sense, that $q_{acc}$ is reached) and rejected if all branches reject. If the machine accepted a language by reaching an accept condition on some branches and rejecting on others, flipping the conditions you end up with a machine which again accepts that language, while you intended to make a machine that rejects it.
Solution 2:
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.
Solution 3:
Classes NP and co-NP are defined using Karp reductions (also known as a polynomial-time many-to-one reductions). A Karp reduction from language $L_1$ to $L_2$ is a polynomial-time computable function $f:\{0,1\}^*\to\{0,1\}^*$ such that $x\in L_1$ if and only if $f(x) \in L_2$. Note that there are no negations in this definition! E.g., if $x\in L_1$ if and only if $f(x) \notin L_2$, then $f$ is not a Karp reduction from $L_1$ to $L_2$.
What you describe in your question is a Cook reduction (also known as a polynomial-time Turing reduction). Every Karp reduction is a Cook reduction but not vice versa. If classes NP and co-NP were defined using Cook reduction then we would have $NP_{Cook}=$ co-NP $_{Cook}$. Note that your argument indeed shows that if $P=NP$ then $P=$ co-NP.