the way i think of conductors is the nuclei (and inner electorns) are "locked" together in a lattice and the the outer electrons flow around this lattice (free electrons means the electrons are bound to a specific atom like with a chemical bond), like a fluid. Also like how water flows to minimise the gravitational energy, the electrons 'flow' to minimise the electrical energy.
So when you have an uncharged conductor with no external electric field, there is no part of it that has a net charge. Because if there was a part of it that was say positively charged then the free electrons would flow towards it, cancelling out the charge, as it would have a lower energy. Alternately if there was a bit that was negatively charged then the free electrons would flow away from that area decreasing the charge, as it would decrease the energy.
Now if the conductor is in an external electric field, like in your image, then the free electrons would flow opposite to the direction of the electric field (the electric field is defined by the force a positive, unit charged particle). so the free electrons flow to one side of the conductor as they are 'pulled' by the electric field.
note: being 'pulled' by a force field is equal to moving to a lowest energy. (warning math) "potential energy"=$-\frac{\Delta force}{\Delta x}$, alternatively $U=-\underline{ \nabla} • \underline{F}$
This would cause a charge difference, like you have in your image, the charge difference between the two faces creates a new electric field, which is equal and oppositely directed to the external electric field, if the fields don't equal then there is still a force on the electrons, so the free electrons flow more until the fields equal, then there is no net force on the electrons.
hopefully that makes helps
