Electron travelling through a step potential from $V_0$ to 0 Most of the time you discuss the step potential case when $E>V_0$, you consider an electron (or a beam of electrons) travelling from a region of space $x<0$ in which $V=0$, to a region $x>0$ in which $V=V_0$, and I think I have understood that case:
the electron has enough energy to cross the step, but due to uncertainity there's a chance of it being pushed back, so the wavefunction modulus will be smaller after the step, and also the wavelength decreases following the rule $λ=\frac{2π\hbar}{\sqrt{2mE}}$ for $x<0$ and $λ'=\frac{2π\hbar}{\sqrt{2m(E-V_0})}$ for $x>0$ 
I can't understand what would happen if we considered the case of an electron travelling from a region with $V=V_0$ to a region with $V=0$:
how could it be pushed back at the step even if the potential energy is decreasing, so the passage should be eased?
I ask this because in an exercise the solution says that the chances of reflection $R_1$ and $R_2 $ are equal in both case of electron "stepping up" and electron "stepping down"



in the bottom drawing the dotted lines represent the reflected waves.
 A: Edit. To begin with, I don't understand the drawings showing incident and reflected waves. They are complex. At the step only imaginary parts vanish, for all wave components.

As to your question, perhaps an analogy could help. Consider a string,
or better two strings of different linear densities tied together. It
you send a wave on the strings, at the junction point there will be a
partial reflection, both if the wave is coming from the lighter string
towards the heavier and the other way around. 
Using the electronics language, it is the impedance mismatch which
causes reflection. Here the place of impedance is taken by the ratio
$\psi'/\psi$. Schrödinger equation is
$$\psi'' + k^2 \psi = 0.$$
If $\psi=e^{ikx}$ then impedance $Z$ is
$$Z = \psi'/\psi = ik$$
and changes at the junction.
The same happens for your problem with a potential step. 
A: To get the physical intuition here, consider a physical apparatus which creates the potential step in the case of an electron. Imagine a series of plane wire grids or meshes, connected by a battery:

The battery gives positive electric potential to region B compared to region A, so electrons (which are negatively charged) will feel a drop down in potential energy when they cross from A to B. That is to say, they experience a large force to the right in the narrow gap between regions A and B where the voltage changes. This is where the step occurs in the potential energy diagram. The electron waves, arriving at this step, experience this strong 'kick', which abruptly forces them to change shape into waves of shorter wavelength. This picture helps one's physical intuition, I think, to see why it is that the electron waves might respond to this situation by being partially reflected. 
Note that it is the abruptness of the kick that causes the reflection. Let $L$ be the width of the narrow region between the two central meshes, where the potential is changing (and so the wavelength is also changing). If the wavelength is small enough, i.e. when $\lambda \ll L$, then the waves can adjust their wavelength continuously as they pass through this region and in this case the reflection tends to zero. When $L \ll \lambda$, on the other hand, the reflection happens (and is calculated in the usual way by stating the solution to Schrodinger's equation and making sure the wavefunction and its gradient are continuous). The "step" potential is an idealisation which takes $L=0$. That is not physically possible; it is really an approximation which is suited to the case $L \ll \lambda$.
A: 
the electron has enough energy to cross the step, but due to uncertainty there's a chance of it being pushed back

$ $

how could it be pushed back at the step even if the potential energy is decreasing, so the passage should be eased?

There is an issue with your interpretation of the wavefunction here. The wavefunction is not the electron. Before we measure the electron to be somewhere, it is not located anywhere. Therefore, we cannot talk about the electron reaching the step and bouncing off or being pushed back.
What we can talk about is how the wavefunction evolves over time according to the Schrodinger equation. Just like how Newton's laws tell us how the classical trajectory of a particle evolves over time, the Schrodinger equation can tell us how the wavefunction evolves.
So what really is being reflected or transmitted across the step is just the wavefunction, which encodes the probability of measuring the electron to be at some position. The step influences these probabilities.
In the case of an initial Gaussian packet approaching the step, there are solutions allowed that have a wave packet traveling in the opposite direction that correspond to measuring the electron to have a momentum in the opposite direction as the initial direction. But we cannot apply classical reasoning, as stated above. The wavefunction reflecting off of the step is not the same thing as the electron reflecting off the step. This is how the wavefunction evolves according to the Schrodinger equation, and so this ends up being a possiblity of our measurements. 
I must also note that in the case of the step up, the wavefunction reflection is not due to uncertainty. For both step scenarios, the cause of wavefunction reflection is due to the same reason as discussed above.
A: One way to think about whole this is by realizing two things:


*

*Schrödinger equation gives the time evolution of a wave function.

*For unitary time independent Hamiltonian (which is the case in your example), taking complex conjugation of Schröginger equation and inverting the direction of the time gives you back the original equation.


So, naively, the complex conjugation of wave function evolves backward in time just the same way the wave function evolves forward in time. If there are multiple wavefunction that mix, then the complex conjugation may result in relative phase differences. However, for one wavefunction, complex conjugation will not change the observables.
Therefore, knowing that wavefunction gets reflected when hitting a higher potential, it is only natural to expect a similar reflection when we go to a lower potential as going from lower to higher potential in forward time is equivalent to going from higher to lower potential in backward time. In fact, realizing this also reveals that the reflection has nothing to do with potential increase. It simply follows from the fact that the potential changes there.
