Free Electron in Current
I'm pretty thankful for Jack's answer – because it explains that you might not want to stick to a model with "separate atoms" and "bouncing" electrons for a metal. So here goes what I'd like you to get the idea of regarding electron movement in a metal:

The moment you realize that these electrons aren't free to move anywhere, you must admit that the word "free electron" isn't 100% accurate.

So far, so good. Hold on, this will hurt just a bit.

The orbits you know are just a model. They don't exist as things with a shape where a "point-shaped" electron circles around. The moment you need to describe electron movement in a metal, that model breaks down, as you've noticed.

Instead, we have to understand that an electron bound to a nucleus only is bound because "fleeing" would require an external impulse, as well as "crashing" into the nucleus. For now, imagine the electron in circular motion (just like a satellite around a planet), and if no external force is applied, it'll stay at that path.

Now, take a step back. You might have heard of Heisenberg's Uncertainty principle – you can't know the exact location of something and its exact impulse at the same time. That's exactly what's happening here – we know the rotational impulse of the electron pretty exactly (because we can calculate how much impulse it needs not to crash nor to flee), and thus, the knowledge of its position must be uncertain to a specific degree.

Hence, an electron like that doesn't actually have a place on the orbit – it has a place probability distribution. It turns out that the probability is an effect (or, rather, an operator applied to) Schrödinger's Equation (for a non-near-speed-of-light single particle), which is

$$i \hbar\frac{\partial}{\partial t} \Psi(\mathbf{r},t) = \left [ \frac{-\hbar^2}{2\mu}\nabla^2 + V(\mathbf{r},t)\right ] \Psi(\mathbf{r},t)$$

(I swear, I'm not trying to scare you – the formula will look far less threatening when you've studied electrical engineering for one and a half years – you'd typically have a course called "solid-state physics / electronics", where this is explained in much more depth and with background, and a lot of mandatory math courses that explain how to deal with this kind of equation, especially with the differential Laplacian operator \$\nabla^2\$. I just need the formula below.)

So, now back from the single electron to the metal:

A metal is composed of an electron lattice – that is, the atoms are arranged in a repetitive pattern. Now, looking at Schrödinger's equation, you'll see a \$V\$ there – that's Potential, and potential is practically "distance to positive charges" for an electron – and since we know the positive charges are in a nice periodic pattern in the metal, \$V\$ is periodic!

Now, what's this \$\Psi\$? It's what we call the position-space wave function. It's the solution for Schrödinger's Equation – the function that makes the "\$=\$" above true!

Now, for a specific, periodic \$V\$, only a specific set of wave functions can exist; we can apply a different operator to the wave function \$\Psi\$ (the Hamiltonian) and get these states; they are the so-called Bloch states. Within these, an electron actually doesn't have a specific "identity" or "place" – it just contributes to the fact that things are periodic.

That's what you mean when you talk of "conduction bands" in metals – states that electrons are a) able to exist and b) are free to move around in.

Now, if you apply an electrical field, which is what you do to, macroscopically, make charges (electrons) flow, you change \$V\$; it's now a sum of a periodic function and a linear function. That leads to a change in the solution for \$\Psi\$ – and macroscopically, this means that electrons move to one end.