No, it doesn't collapse to an eigenstate. Collapse to an eigenstate is a picture of an ideal measurement. In general the final state will not be describable by a wave function, because it's not a pure state, it is instead a mixed state. See this question, which is about inexact measurements.
Position eigenstate in position representation is $\langle x_{}|x_0\rangle=\delta(x-x_0)$. This gives the following in the momentum representation: $\langle p_{}|x_0\rangle=e^{\frac{i}{\hbar}px}$. For this function probability density is constant, thus its expectation value is undefined (one can't find a center of infinite line). Similarly, for free particle expectation value of energy will also be undefined. This is because such state is an abstraction, a useful mathematical tool. Of course, such states can't be prepared in real experiment, but one can come very close to it, e.g. shoot an electron at a tiny slit and observe state of the electron at the very exit of that slit.
As to finding expectation value of energy in position eigenstate, first mistake which you make using the formula $\overline E=\langle x|\hat H|x\rangle$ is forgetting to normalize the eigenvector. But position operator has continuous spectrum, which makes all its eigenvectors unnormalizable (i.e. if you try to normalize them, you'll get null vector, which is meaningless as a state). Thus you can't directly find expectation value of energy in position eigenstate.
I will answer this part
In addition, we know that the Hamiltonian represents the sum of kinetic and potential energy in a system.However, I'm not quite sure why, intuitively, the time dependent version of the Schrodinger equation becomes Hψ=iℏ ∂/∂t ψ(r,t).
Quantum mechanics was developed slowly, because experiments showed that light came in quanta from the hydrogen atom. At that time they were still thinking classically, and Bohr developed a model of an electron rotating around a proton for the hydrogen atom, similar to the way the moon rotates around the earth. BUT there was a problem for this. In classical electricity and magnetism the electron would not stay in an orbit but would lose energy and fall on the proton.
Bohr postulated that it was a standing wave, and postulated only certain orbits ; electrons could fall from one to the other emitting a photon of energy hnu ( nu the frequency). That the photon's energy came as hnu was known from the photoelectric effect and from black body radiation. The model then explained the spectrum of hydrogen which had been fitted with a series.
This is how the h enters the game. Because the model has to take into account that an electron changing energy levels will release from the system energy proportional to h*nu.
The Schrodinger equation gives the same series as a solution to the hydrogen problem, but now it comes as a theory which is much more general. h necessarily has to play its role.
Quantum mechanics has a number of postulates.
Associated with any particle moving in a conservative field of force is a wave function which determines everything that can be known about the system.
With every physical observable q there is associated an operator Q, which when operating upon the wavefunction associated with a definite value of that observable will yield that value times the wavefunction.
Any operator Q associated with a physically measurable property q will be Hermitian.
The set of eigenfunctions of operator Q will form a complete set of linearly independent functions.
For a system described by a given wavefunction, the expectation value of any property q can be found by performing the expectation value integral with respect to that wavefunction.
The time evolution of the wavefunction is given by the time dependent Schrodinger equation.
Number 2) of these postulates is what relates to your question.
where does the iℏ come from? why does the sum of kinetic and potential energy equal to that?
The h bar comes so that the dimensions and the energy of the photon comes out correctly as h*nu. The complex "i" so that the equation has the form that will give the appropriate solutions.
The operator for the time dependent Hamiltonian is $i\hbar$$\partial$/$\partial$$t$
So the $H$$\psi$=$i\hbar$ $\partial$/$\partial$$t$ is an identity, used to solve for a time dependent psi:
The formalism developed by trial and error in the beginning, fitting the models to the data and then using the models to predict further behaviors.The successful fitting of the same spectral series as the Bohr model led to the development of quantum mechanics, rather than the theory coming first and then looking at the data.
The real answer is that this mathematical formulation fits the data and has great predictive power proven over and over again.
Best Answer
If an ideal position measurement is made and the particle if found to be at $\mathbf{x}$, and then another ideal position measurement is made immediately after the first, the result of the second measurement is certain to be $\mathbf{x}$. So uncertainty in the result of this second position measurement does indeed vanish.
However, if the second measurement is instead an ideal momentum measurement, any result is equally probable, i.e., there is 'infinite' uncertainty in momentum.
So it isn't true that any uncertainty should vanish.