States with defined energy and time evolution Consider the following simple problem:
We have a step potential:
$$V=V_0\Theta (x)$$
so the Hamiltonian is:
$$H=\frac{p^2}{2m}+V_0\Theta(x)$$
and we want to find the eigenfunctions of the Hamiltonian $\psi _E$, so the states with defined energy for our system. This is easy enough to understand and solve.

Since $H$ is not a function of $t$ the time evolution operator $S(t,t_0)$ acting on an eigenfunction of $H$ with energy $E$ has the following structure:
$$S(t,t_0)=\exp{\left[\frac{1}{i\hbar}E(t-t_0)\right]}$$
Wonderful! Now we can use the time independent Schrodinger equation:
$$H|\psi _E\rangle = E |\psi _E \rangle \ \Rightarrow \ \langle x |H|\psi _E\rangle = E \psi _E (x)$$
and so we get two different differential equations: one for $x<0$ and one for $x \geq 0$. Form here, with some calculation, we get plane waves on both sides if $E>V_0$ and plane waves on the left side and decreasing exponential in the right side if $E<V_0$. At last if we like we can specify the time evolution, given by $S$, to write $\psi(x,t)$ instead of $\psi(x)$:
$$\psi _E(x,t)=\psi _E(x)\exp{\left[\frac{1}{i\hbar}E(t-t_0)\right]}$$

Wonderful. I have no problem with this.
But now suppose that our objective changes: we now want to find, for the same system, what happens when a plane wave (so a particle with wave function that is a plane wave) coming from the left side impacts our system.
I feel like I don't know how to solve this problem: this problem seems like a scattering problem, or a time evolution problem where the function is not an eigenfunction of the system.
But I have seen that the way to solve this problem is practically identical to the way we solved the first question: to solve this last question we simply find the eigenfunction of the system, as if the objective was to find them, and then, in the case of $E>V_0$, we simply remove, the plane wave on the right side with $k<0$, and this should represent the fact that the wave comes from the left side.
I know how to solve this last question, but I don't understand why this procedure work! For example: why the particle coming from the left, after impacting the barrier, stays into an eigenstate with defined energy? What theorem ensures this? Couldn't the impact put the particle into a superposition of energy? Remember that the particle has wavefunction that is eigenfunction of $H$ on the left side, but not in all space.
Even worse: the procedure that I cited seem to work almost for all potentials: step potential, well potential, Dirac's Delta potential, and so on. So there must be something essential about quantum particles "dynamics" that I am missing here.
TL;DR: What do the eigenfunction of $H$ have to do with a scattering problem for a plane wave?
(Bonus question: what if the particle impacting from the left side has wavefunction that is not a plane wave? What if, for example, we have a gaussian wave packet coming from the left and impacting our potential?)
 A: In $1D$, when you are looking for the stationary states (ie the eigenstates of the Hamiltonian), you are trying to solve the second order linear differential equation :
$$\frac{-\hbar ^2}{2m}\frac{\text d^2\psi}{\text dx^2}+ V(x)\psi(x) = E\psi(x)$$
(and to find solutions which can be properly normalized).
If the potential function $V(x)$ satisfies :
$$V(x) = \left\{ \begin{array}{cl}
V_- & \text{for } x\text{ near } - \infty \\
V_+ & \text{for } x \text{ near } + \infty
\end{array}\right.$$
and you are looking for solutions with $E\geq \max (V_-, V_+)$, then you see that the solution is given, near $\pm\infty$ by :
$$\psi(x)  = A_\pm e^{ik_\pm x} + B_{\pm}e^{-ik_\pm x}$$
where :
$$k_{\pm} = \sqrt{\frac{2m(E - V_\pm)}{\hbar^2}}$$
More precisely, what this means is that, by setting the constant $A_-$ and $B_-$ (ie the form of the solution near $-\infty$), you can extend it up to $+\infty$ (as this is a second order linear differential equation) where it will have also a simple expression. Put in another way, of the $4$ constants $A_\pm$, $B_\pm$, only two are free parameters for the solution.
One of this state corresponds physically to a situation where a plane wave with amplitude $A_-$ and wavevector $k_-$ is incident from $-\infty$ and a plane wave with amplitude $B_+$ and wavevector $-k_+$ is incident from $+\infty$. If you want to find the state which corresponds to a plane wave  with amplitude $1$ incident from $-\infty$ (implying that no wave is incident from $+\infty$), you have to find the solution with $A_- = 1$ and $B_+ = 0$.
