Unfortunately, "in general" has both a colloquial meaning in English and a mathematical meaning.
The colloquial meaning is the same as "usually". For example, "I generally wake up at 6 am, but sometimes I oversleep."
The mathematical meaning is "true for all instances", as in "the eigenvalue of a Hermitian operator will be real in general", since all eigenvalues of Hermitian operators are real.
When someone says "operators are generally not commutative", it is somewhat ambiguous which meaning they intend. It is most likely the colloquial meaning, so they meant "most of the time, operators don't commute." If they intended the mathematical meaning, they meant "operators do not always commute."
I don't understand your follow-up about adjoints. Certainly, if operators do not commute in general, neither do adjoints of operators (after all, the set of all adjoints is the same as the set of all operators).
If an operator's adjoint is equal to its complex conjugate, it means the operator is symmetric. To see this, simply write the relation in component form. If we represent the components of the original operator by $A_{ij}$, the equation you gave is simply
$$A_{ji}^* = A_{ij}^*$$
which implies
$$A_{ji} = A_{ij}$$
so the operator is symmetric.
Your last question does not make sense to me. The equation $\left[A,B\right] = AB - BA$ is the definition of the commutator. Perhaps you made a typo in your math typesetting? If you are asking whether it is true that
$$\left[A^\dagger, B \right] = A^\dagger B - B^\dagger A$$
then no, that is not correct. Simply by substituting $A = A^\dagger$ in the original expression for the commutator, we get
$$\left[A^\dagger, B \right] = A^\dagger B - B A^\dagger$$
I) First of all, one should never use the Dirac bra-ket notation (in its ultimate version where an operator acts to the right on kets and to the left on bras) to consider the definition of adjointness, since the notation was designed to make the adjointness property look like a mathematical triviality, which it is not. See also this Phys.SE post.
II) OP's question(v1) about the existence of the adjoint of an antilinear operator is an interesting mathematical question, which is rarely treated in textbooks, because they usually start by assuming that operators are $\mathbb{C}$-linear.
III) Let us next recall the mathematical definition of the adjoint of a linear operator. Let there be a Hilbert space $H$ over a field $\mathbb{F}$, which in principle could be either real or complex numbers, $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$. Of course in quantum mechanics, $\mathbb{F}=\mathbb{C}$. In the complex case, we will use the standard physicist's convention that the inner product/sequilinear form $\langle \cdot | \cdot \rangle$ is conjugated $\mathbb{C}$-linear in the first entry, and $\mathbb{C}$-linear in the second entry.
Recall Riesz' representation theorem: For each continuous $\mathbb{F}$-linear functional $f: H \to \mathbb{F}$ there exists a unique vector $u\in H$ such that
$$\tag{1} f(\cdot)~=~\langle u | \cdot \rangle.$$
Let $A:H\to H$ be a continuous$^1$ $\mathbb{F}$-linear operator. Let $v\in H$ be a vector. Consider the continuous $\mathbb{F}$-linear functional
$$\tag{2} f(\cdot)~=~\langle v | A(\cdot) \rangle.$$
The value $A^{\dagger}v\in H$ of the adjoint operator $A^{\dagger}$ at the vector $v\in H$ is by definition the unique vector $u\in H$, guaranteed by Riesz' representation theorem, such that
$$\tag{3} f(\cdot)~=~\langle u | \cdot \rangle.$$
In other words,
$$\tag{4} \langle A^{\dagger}v | w \rangle~=~\langle u | w \rangle~=~f(w)=\langle v | Aw \rangle. $$
It is straightforward to check that the adjoint operator $A^{\dagger}:H\to H$ defined this way becomes an $\mathbb{F}$-linear operator as well.
IV) Finally, let us return to OP's question and consider the definition of the adjoint of an antilinear operator. The definition will rely on the complex version of Riesz' representation theorem. Let $H$ be given a complex Hilbert space, and let $A:H\to H$ be an antilinear continuous operator. In this case, the above equations (2) and (4) should be replaced with
$$\tag{2'} f(\cdot)~=~\overline{\langle v | A(\cdot) \rangle},$$
and
$$\tag{4'} \langle A^{\dagger}v | w \rangle~=~\langle u | w \rangle~=~f(w)=\overline{\langle v | Aw \rangle}, $$
respectively. Note that $f$ is a $\mathbb{C}$-linear functional.
It is straightforward to check that the adjoint operator $A^{\dagger}:H\to H$ defined this way becomes an antilinear operator as well.
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$^{1}$We will ignore subtleties with discontinuous/unbounded operators, domains, selfadjoint extensions, etc., in this answer.
Best Answer
There is no unique canonical notion of complex conjugation $C:H\to H$ of vectors in an abstract complex Hilbert space $H$. However, given a notion of complex conjugation $C:H\to H$, it is naturally to demand that it is an antiunitary map $$\forall v,w\in H:~~\langle C(v) | C(w)\rangle~=~\overline{\langle v | w\rangle}.\tag{1}$$ (This is e.g. the case for the Hilbert space $L^2(\mathbb{R}^3)$ equipped with the standard sesquilinear form $\langle \cdot | \cdot\rangle$ and complex conjugate.)
Since $C$ is an involution, eq. (1) is equivalent to the definition that the antilinear map $C$ is "Hermitian" $$\forall v,w\in H:~~\langle C(v) | w\rangle~=~\overline{\langle v | C(w)\rangle},\tag{2}$$ in the sense that $C^{\dagger}=C$, cf. e.g. this Phys.SE post.