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").
The eigenstates of the operator $N = a^\dagger a$ can be labeled by their eigenvalues, i.e. $N \phi_n = n \phi_n$, where $n$ is an integer. Note that
$$n = \langle \phi_n,N\phi_n\rangle = \underbrace{\langle\phi_n,a^\dagger a \phi_n\rangle = \langle a \phi_n,a\phi_n\rangle}_{a \text{ and } a^\dagger\text{ are mutually adjoint}} = \Vert a \phi_n\Vert^2 \geq 0 $$
In particular, we have that
$$ 0 = \Vert a\phi_0 \Vert^2$$
Since the only element of a Hilbert space with zero norm is the zero vector itself, we have that
$$a \phi_0 = 0$$
Best Answer
No, that relationship is incorrect. If you start with $$ \hat{a}|n\rangle = \sqrt{n}|n-1\rangle $$ and take the conjugate, what you get is $$ \langle n|\hat a^\dagger = \sqrt{n}\langle n-1|. $$ To get $\langle n|\hat a$, you need to start instead with an annihilation operator, $$ \hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle, $$ so that when you take the conjugate you get $$ \langle n|\hat a = \sqrt{n+1}\langle n+1|. $$