I'm answering only part of your question, please bear in mind that some of the topics that you pointed out are doubts I have myself.
Concerning the comment from Feynman, I usually don't like the authority arguments. If your only reason to believe something is that someone said it, it's not a good reason. Just as some examples, Newton believed that light had no wave-like behaviour, which is simply wrong.
About the variational formulation of Schrödinger Equation. If by variational you mean, having a Lagrangian, the following Lagrangian does the job:
$
\mathcal L = \frac{i\hbar}{2}\left(\psi^*\partial_t \psi - \psi \partial_t \psi^*\right) - \frac{\hbar^2}{2m}\left(\nabla \psi\right)\cdot \left(\nabla \psi^*\right) - V\psi^*\psi
$
With the assumption that you have a complex classical scalar field, $\psi$, and that you can calculate the Euler-Lagrange equation separately for $\psi$ and $\psi^*$. As with anything using the extremum action principle, you really have to guess which your Lagrangian is based in Symmetry principles + some guideline for your problem, here is no different.
I don' t know how did Schrödinger made the original derivation, but if you make a slight change of variables in the above lagrangian, you get very close to what you have written above.
The trick is writing $\psi = \sqrt n e^{iS/\hbar}$, where $n$ is the probability density and $S$ is, essentially the phase of the wave-function. If you have trouble with the derivation, I can help you latter.
Again about Feynman's quote. It's possible to arrive at Schrödinger equation without passing through an arbitrary heuristic procedure, that way is developed by Ballentine's book. You still have to postulate where you live, if you consider that arbitrary or not, it is completely up to you.
The other point that it is possible to consistently and rigorously construct a quantum theory from any classical mechanical theory, using quantization by deformation. That's why I think it's a bit false that "It's not possible to derive it from anything you know. It came out of the mind of Schrödinger", as Feynman, and many other great names, said.
The "independent" in "time-independent Schrödinger equation" doesn't mean that the wavefunction $\psi(x,t)$ is independent of time, but that the quantum state it defines doesn't change with time.
Since $\psi(x)$ and $\mathrm{e}^{\mathrm{i}\phi}\psi(x)$ for any $\phi\in\mathbb{R}$ define the same quantum state, this does not imply $\partial_t\psi(x,t) = 0$. Indeed, as the solution shows, the time dependence $\mathrm{e}^{\mathrm{i}Et}$ is precisely the kind of dependence that is allowed.
Best Answer
The point made by Zweibach in those notes is true even for time-independent Hamiltonians.
Let $\psi:\mathbb R\rightarrow \mathscr H$ be a trajectory through the Hilbert space $\mathscr H$, where $\psi(t)$ is understood to be the state vector of the system at time $t$. The Schrodinger equation is $$i\hbar \psi'(t) = H(t) \psi(t)\tag{1}$$
Let us assume that for each $t$, $\psi(t)$ is an eigenvector of $H(t)$ with eigenvalue $E(t)$. That is, $$H (t)\psi(t) = E(t)\psi(t)\tag{2}$$
Zweibach's point is that it does not follow that $\psi(t)$ is a solution to $(1)$. Plugging $(2)$ into $(1)$ yields $$i\hbar \psi'(t) = E(t) \psi(t) \implies \psi'(t) = -\frac{i E(t)}{\hbar} \psi(t)$$
Obviously this equation is not satisfied by arbitrary trajectories $\psi$; even if the Hamiltonian is time-independent and $E(t) \equiv E_0$ is constant, in order for $\psi$ to be a solution to $(1)$ we must have that $$\psi(t) = e^{-iE_0 t/\hbar} \psi(0)$$
As an explicit example, consider the elementary particle in a box of length $L$. The ground state eigenvector for this system is $\psi_0 = \sqrt{2/L} \sin\big(\pi x/L\big)$ with eigenvalue $E_0 = \pi^2\hbar^2/2mL^2$. If we let $\psi(t)=\psi_0$, we see that for each $t$, $\psi(t)$ is an instantaneous eigenvector of the Hamiltonian (meaning that $H\psi(t) = E_0 \psi(t)$); however, it is also obvious that $\psi(t)$ does not solve the (time-dependent) Schrodinger equation $(1)$.