Current flow between metal and piece of pure silicon
Good question - there's a lot to unpack here, it's not as straightforward as it appears! Before answering your questions, a little bit of quantum mechanics is needed to understand it.
First:

My understanding of the metal wire is that: It has a very low bandgap energy and thus the covalent bonds in the valence band that the atoms of the metal form with each other break extremely easily and hence we have a lot of electrons in the conduction band, available to flow, just waiting for a voltage to be applied to start moving!

I'd correct that a little: metals don't have a bandgap energy at all, at least not as we'd think compared to semiconductors (they have bands, just not near the Fermi level). The Fermi level (the energy level where there's a 50% chance of an electron state being occupied) is in the middle of a band for a metal, which is why there are so many electrons available to carry current. In an undoped semiconductor, it's in the middle of a band gap - far away from any electron state, so there are not many carriers available (unless you make it very hot). When you dope the semiconductor, you move that Fermi level closer to a band edge, increasing the likelihood of an electron in the conduction band, or the lack of an electron (hole) in the valence bad. If you highly dope a semiconductor, you can drive the Fermi level into the conduction band - making it very metal-like.
Second: You are asking about the resistance of a material, and trying to compare the resistance of metal to semiconductor. The resistance is derived from a term called &quot;resistivity&quot; times material length divided by area. You are correct that resistivity (\$\rho\$) is a function of carrier density, which is higher in a metal than an undoped semiconductor. But other things also matter, such as temperature and carrier mobility. Carrier mobility in turn depends on things like the quantum mechanical effective mass of a carrier in the semiconductor, and how scattering the semicondcutor is. The effective mass is basically &quot;how much mass does an electron appear to have based on the relationship between its wavefunction's momentum in a crystal lattice and its energy?&quot; Holes tend to be &quot;heavier&quot; than electrons.
Ok, with that background we can get to your questions:

My question is understanding how this current is limited from a atomical level?

Assuming you care about the atomic level (and can ignore material area and length), current is limited by the availability of carriers (e.g., semiconductor doping, semiconductor temperature), how scattering the material is (semiconductor crystal purity and temperature of a metal), what the effective mass of a carrier is (an electron in a metal typically weighs less than an electron in a semiconductor conduction band typically weighs less than a hole in a semiconductor valence band).

The metal has excess free electrons, right?

If a metal is electrically neutral, it has just the right amount of free electrons. That is - it has as many free electrons as there are + metal ions. If it had excess, it would have a net negative charge.

Why can't these excess free electrons travel through the lattice?

They can and do. The problem is getting in to the lattice. The electrons in the metal need a state in the semiconductor for them to be able to transition in to. They either have to quantum mechanically tunnel through the bandgap barrier (e.g., Schottky contact), or the semiconductor has to be doped such that the electrons in the metal have a suitable state without tunneling (e.g., ohmic contact). See: https://en.wikipedia.org/wiki/Metal%E2%80%93semiconductor_junction

Why do we have to rely on the limited number of free electrons in the silicon piece?

This is the same question as &quot;how come a resistor limits the current through a diode?&quot; Since the same current has to flow through all elements in series, one highly resistive element will &quot;limit&quot; the current through all the other elements in series. So the same number of electrons will go through both the semiconductor and the metal, it's just that the semiconductor will have a bigger voltage drop and effectively limit the total current through the entire series (assuming everything is set up correctly so that current can flow between the metal and semiconductor properly in the first place)