The wording used in your textbook was sloppy.
$A$ acts as $A^*$ on a bra, as $\langle u\rvert A\lvert v\rangle:=\langle u\lvert Av\rangle~$ is the same as $\langle u\rvert A\lvert v\rangle=\langle A^*u\lvert v\rangle~$, by definition of the adjoint. The latter formula also shows that $\langle A^*u\rvert=\langle u\rvert A$.
Everything becomes very simple in linear algebra terms when interpreting a ket as a colum vector, the corresponding bra as the conjugate transposed row vector, an operator as a square matrix, and the adjoint as the conjugate transpose. This is indeed the special case when the Hilbert space is $C^n$.
Recall the definition of an inner product space, a vector space equipped with a map $n$ that returns a scalar for any pair of vectors. In quantum mechanics, the Hilbert space of states for any quantum system is always an inner product space with the interpretation you noted above: The inner product of a state $A$ on a state $B$, $\langle A|B\rangle$, is the probability amplitude of measuring the system to be in state $A$ given you have knowledge the system is in state $B$.
Rather than considering the set of vectors and a two vector map $n(w,v)$, we can take the perspective instead that there is a dual space. Physical state vectors are "kets" if you will, but there exists another vector space, called the dual space, which is filled with "bras". These bras form the space of linear maps from the kets to complex numbers (the scalars in quantum mechanics).
Notice that there is a one-to-one correspondence between the bras and kets since, by linearity, operators in the dual space are fully defined by their action on a complete set of basis vectors in the ket space. That means that we can have a notion of taking a ket to a bra and back, normally we denote that by $\dagger$, but also sometimes by $*$ or a bar.
And, again using the correspondence, we parameterize the space of bras by objects, $\langle x |$, with the property $\langle x | y \rangle = \delta_{xy}$. Immediately we can work out the transformation properties for an orthonormal basis of bras by considering the action of a properly normalized bra on a properly normalized ket
$$|x\rangle = a_1 |a_1\rangle + a_2 |a_2\rangle.$$
The dual vector must be given by
$$\langle x | = \bar{a}_1 \langle a_1 | + \bar{a}_2 \langle a_2| $$
if we are to have that
$$\begin{align}\langle x|x\rangle &= \bar{a_1} a_1 \langle a_1|a_1\rangle + \bar{a_2} a_2 \langle a_2|a_2\rangle + \bar{a_1} a_2 \langle a_1|a_2\rangle + \bar{a_2} a_1 \langle a_2|a_1\rangle \\
&= \bar{a_1} a_1 \langle a_1|a_1\rangle + \bar{a_2} a_2 \langle a_2|a_2\rangle + 0 + 0 \\
&= 1,\end{align}$$
which is required by orthonormality.
Best Answer
I'm not totally sure what the question is. If you hand me a vector $f$ I can hand you a continuous linear functional $F : \phi \mapsto \langle f,\phi\rangle$. If you hand me a continuous linear functional $F$, then I can hand you a vector $f = \sum_i \underbrace{\overline{F(\varphi_i)} }_{\in \mathbb C}\ \varphi_i$, where $\{\varphi_i\}$ is any orthonormal basis of the Hilbert space and the line denotes complex conjugation.
Note that the fact that this indeed a vector in the Hilbert space is trivial for finite-dimensional spaces, and follows from the fact that all continuous linear functionals are bounded for infinite-dimensional spaces.
One can show without too much difficulty that this correspondence amounts to an isomorphism - a one-to-one pairing between vectors and continuous linear functionals. This pairing is antilinear in the following sense: if $f$ and $g$ are vectors, $F$ and $G$ the corresponding continuous linear functionals, and $\alpha\in \mathbb C$ then $$f + g \iff F + G \qquad \text{and} \qquad \alpha f \iff \overline\alpha F$$
Both statements can be proven straightforwardly from the pairing defined above.
Yes, that's right. Indeed, we can go the other way: given a vector $f$ we can produce a continuous, antilinear functional $\widetilde F:\phi \mapsto \langle \phi,f\rangle$, and given a continuous antilinear functional $\widetilde F$ we can produce a vector $f = \sum_i \widetilde F(\varphi_i) \varphi_i$. This correspondence is then linear in the sense that $$f + g \iff \widetilde F + \widetilde G\qquad \text{and}\qquad \alpha f \iff \alpha \widetilde F$$
Having built up all this structure, given a vector $f$ we use the shorthand $F \equiv \langle f|$ for its linear functional partner, and $\widetilde F \equiv |f\rangle$ for its antilinear functional partner, while defining $\langle f|g\rangle := \langle f,g\rangle$. This provides the rigorous underpinning of the bra-ket notation which is in universal use among working physicists.