I'm going to assume that Omer is specifically asking why the centre of mass is at the focus (well, one of the foci) of the orbits. Omer, if this isn't what you meant please ignore what follows because it's completely irrelevant!
If you have a body moving in a central field (i.e. the force is always pointing towards the centre), and the field is inversely proportional to the square of the distance from body to the centre, then the orbit is an ellipse with the centre at one of the foci. For now let's just assume this and we can come back to prove it later.
So if we can show that both of the bodies feel a central inverse square force, with the COM at the centre, this guarantees the orbits will be ellipses with a focus at the COM. Given that the force is due to the two bodies attracting each other, and that both bodies are orbiting around, it may seem a bit odd that each body just feels a central inverse square force, but actually this is easy to show.
The picture shows the two masses and the COM. I haven't shown the velocities because it doesn't matter what they are. For now let's just consider $m_1$ and calculate the force on it. By Newton's law this is simply:
$$ F_1 = \frac{Gm_1m_2}{(r_1 + r_2)^2}$$
First is this force central? We know the centre of mass doesn't move. For two bodies this seems obvious to me, but in any case dmckee proved it in his answer. If the COM doesn't move it must lie on the line joining the two mases, otherwise there'd be a net force on it. So the force $F_1$ must always point towards the COM i.e. the force is central.
Second is this an inverse square law force i.e. is $F_1 \propto 1/r_1^2$? Well the definition of the centre of mass is that:
$$m_1r_1 = m_2r_2 $$
or
$$ r_2 = r_1 \frac{m_1}{m_2} $$
If we substitute for $r_2$ in the expression for $F_1$ we get:
$$ F_1 = \frac{Gm_1m_2}{(r_1 + r_1(m_1/m_2))^2}$$
or with a quick rearrangement:
$$ F_1 = \frac{1}{(1 + m_1/m_2)^2} \frac{Gm_1m_2}{r_1^2}$$
and this shows that $F_1$ is inversely proportional to $r_1^2$. I won't work through it, but it should be fairly obvious that exactly the same reasoning applies to $F_2$ so:
$$ F_2 = \frac{1}{(1 + m_2/m_1)^2} \frac{Gm_1m_2}{r_2^2}$$
This is the key result. Even though the two bodies are whizzing around each other, each body just behaves as if it were in a static gravity field, but the strength of the field is reduced by a factor of $(1 + m_1/m_2)^2$ for $m_1$ or $(1 + m_2/m_1)^2$ for $m_2$. This applies to all two body systems, even such unequal ones as the Sun and the Earth (ignoring perturbations from Jupiter etc).
I did start by assuming that a body in a central gravity field orbits in an ellipse with the foci at the centre, but I'm going to wimp out of proving this since it would double the length of this answer and you'd all go to sleep. The proof is easily Googled.
NB this only applies to two body systems. For three or more body systems the orbits are generally not ellipses with the centre of mass at the focus.
There are several inaccuracies in your post.
First - you don't need equal mass (or one much smaller than the other) for a circular orbit: you need the least amount of energy for a given angular momentum. You can have unequal masses in circular orbit (about their center of gravity).
Second - while the force of gravity is pulling towards the focus of the ellipse, the direction of the orbit need not be perpendicular to the radial vector. When it is not, then part of the force will "slow down" or "speed up" the object in orbit, and part of it will "bend" the direction:
In this case, the red component is the one that is providing centripetal force, and the green component is accelerating the particle in orbit.
Best Answer
The second (empty) focus is relevant in the theory of tides. In an elliptical orbit, the line joining the planet and the empty focus rotates at the same frequency as the mean motion of the planet; therefore, if spin rotation period is equal to the orbital period (the planet is locked in a synchronous rotation), the planet rotates with one face pointing to the empty focus.
Importantly, a tidal bulge will try to point to the massive object (the occupied focus) while the planet itself will be pointing to the empty focus, causing a "librational tide".