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 $C^n$.
Bra-ket notation is just a useful short hand for some well-defined objects in functional analysis (or linear algebra if you work in finite dimensions).
To understand what is allowed and what isn't, you would better know what those concepts are, so let's quickly recap:
- A ket $|\psi\rangle$ is just a vector in some Hilbert space $\mathcal{H}$.
- A bra $\langle \psi |$ is an object in the dual space of that Hilbert space (usually denoted $\mathcal{H}^*$), which is the space of all linear functionals on the Hilbert space. This implies that $\langle \psi |$ is a linear function, where the variable runs over all possible $|\phi\rangle\in \mathcal{H}$ and the result is a number, i.e. $\langle \psi|$ is a function $\mathcal{H}\to \mathbb{C}$.
- The dual space of a Hilbert space is itself a vector space, hence it makes sense to add bras and multiply them with scalars.
- Now, given the scalar product of the Hilbert space $\langle \cdot, \cdot \rangle$, you can easily construct such functionals: For every $|\phi\rangle\in \mathcal{H}$, the map $|\psi\rangle \mapsto \langle |\phi\rangle,|\psi\rangle\rangle$ is a linear functional. Identifying this linear functional with $\langle \phi |$ lets you write $\langle \phi |\psi \rangle$, which hints at the fact that you used the scalar product.
- According to some theorem, that's all there is to the dual space, hence the bra-ket notation captures the Hilbert space, its dual and the natural duality between the two (the inner product).
- An operator is now a linear function $A:\mathcal{H}\to \mathcal{H}$. This means that it takes vectors to vectors.
- This explains why what you write as "prohibited" is at first glance ill-defined. However, beware: some of the "prohibited" notions are actually all over other texts. For instance, $|\psi\rangle|\phi\rangle$ is often used as a shorthand for $|\psi\rangle \otimes |\phi\rangle$, but clearly, this has to be defined!
Now back to your question. The main problem is, as you say, that the differentiation is not an operator in the sense above. Why? Because it doesn't really act on your Hilbert space. $|\psi(0)\rangle\in \mathcal{H}$. Also $|\psi(t)\rangle\in \mathcal{H}$ for any fixed $t$, but the Hilbert space doesn't know anything about $t$. $|\psi(t)\rangle$ is in fact a function from some time interval $[a,b]$ to the Hilbert space $\mathcal{H}$. And it's on that parameter that the differential acts.
You can also see this with matrices: Given $\mathcal{H}=\mathbb{R}^2$ as your Hilbert space, we have for example a vector $v\in \mathcal{H}$ which is just a column vector. Also, an operator is a two-by-two matrix (that's the linear functions $\mathbb{R}^2\to\mathbb{R}^2$. Let's pick an operator and call it $A$. Then we can write:
$$ v=\begin{pmatrix}{} v_1 \\ v_2 \end{pmatrix}\quad \mathrm{and}\quad A=\begin{pmatrix}{} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} $$
But what about the $t$? Well, clearly, this means that all coordinates are dependent on $t$. You can even have operators that depend on $t$ (see for instance the Heisenberg picture):
$$ v(t)=\begin{pmatrix}{} v_1(t) \\ v_2(t) \end{pmatrix}\quad \mathrm{and}\quad A(t)=\begin{pmatrix}{} a_{11}(t) & a_{12}(t) \\ a_{21}(t) & a_{22}(t) \end{pmatrix} $$
Now if the time differentiation was an operator, you'd have to be able to write it as a two-by-two matrix. Can you? Obviously not. And the reason is that time is just a parameter. It doesn't really belong to the Hilbert space structure (it's not a dimension or anything).
From this picture, you can also see why what you wrote down is valid. How would you differentiate a scalar product of $v,w\in \mathbb{R}^2$? Well, let's write it out:
$$ \frac{\mathrm{d}}{\mathrm{d}t} (v(t)\cdot w(t)) = \frac{\mathrm{d}}{\mathrm{d}t}(v_1(t)w_1(t)+v_2(t)w_2(t))$$
Now you just have a bunch of scalar functions and you can use the usual rules of differentiation and obtain the result.
Can you make time an operator? That's an entirely different question (the answer is "not really").
Best Answer
Here's how you get from one to the other. Let me take the case of a particle moving in one dimension.
In this case, we assume that there exist vectors $|x\rangle$ which form a "dirac-normalized" basis (the position basis) for the Hilbert space in the sense that their inner products satisfy $$ \langle x|x'\rangle =\delta(x-x') $$ Note, as an aside, that these vectors are not normalizable in the standard sense, and therefore they do not strictly speaking belong to the Hilbert space.
Next, for each $|\psi\rangle$ in the Hilbert space, we define the position basis wavefunction $\psi$ corresponding to the state $|\psi\rangle$ as $$ \psi(x) = \langle x|\psi\rangle $$ So really, the value $\psi(x)$ of the position basis wavefunction $\psi$ at a point $x$ can simply be thought of as the basis component of $|\psi\rangle$ in the direction of $|x\rangle$ just as in the finite-dimensional case where one can find the component of a vector $|\psi\rangle$ along a basis vector $|e_i\rangle$ simply by taking the inner product $\langle e_i|\psi\rangle$.