This video shows a telescope in a Halo or Lissajous orbit near the Sun-Earth L2 point. By moving slowly and tracing its path in both inertial and rotating (synodic) frames, you can see that the orbit is primarily around the sun, and the "orbit around" the libration point is a construct that comes from doing the math or thinking in a rotating frame:
Turn your audio down or off first!
Here's a simplified way to look at the situation:
Orbits are the paths that bodies follow in response to forces.
Forces are the gradients of potentials.
For a small planet around a big star, the potential is
$$\phi(r) \ = \ GM\frac{1}{r}$$
so the force felt by the planet is
$$F(r) \ = \ -GM\frac{1}{r^2}$$
pointed towards the star. When you read about Keplerian orbits, you're reading about only this situation, or a slightly modified version where the planet is bigger and you treat the two as orbiting around their center of mass.
In vector notation, you can change the force to
$$\mathbf{F}(\mathbf{r}) \ = \ -GM\frac{\mathbf{r}}{r^3}$$
If you calculate the trajectory of bodies with various starting conditions in this force field, they will all be Keplerian orbits.
In a three-body problem, and lets stick with the simple "circular restricted three body problem (CR3BP)" where there are two main bodies (a sun and a large planet for example) in circular orbits around their center of mass, the potential field and resulting force felt by the third body (so small you can treat its mass as zero) is
$$\phi(\mathbf{r}) \ = \ GM_1\frac{1}{r_1}+GM_2\frac{1}{r_2} $$
$$\mathbf{F}(\mathbf{r}) \ = \ -GM_1\frac{\mathbf{r_1}}{r_1^3}-GM_2\frac{\mathbf{r_2}}{r_2^3}$$
where $\mathbf{r_1}$ and $\mathbf{r_2}$ are the vectors which point from each of the two main bodies to the third body. Now the shape of the potential is constantly changing. Since the problem is kept simple by requiring the two bodies to be in circular orbits around each other, the potential field is just rotating.
This means that the force field is also rotating, and the orbit of the third body will be determined by its initial velocity and the acceleration due to that rotating force field.
The resulting orbit can be all kinds of crazy things, but it won't be a nice Keplerian orbit, and so none of that math applies.
To mathematically solve for orbits in this problem, mathematicians and orbital mechanics often switch to doing their math in a rotating frame, one that rotates with the two main bodies. In this frame they create a pseudopotential with a term called "centrifugal potential" which is really just a way to pretend the frame is not rotating.
In other words, in the rotating frame, along with the real gravitational potential from the two bodies (above) which is not affected by the rotation, and who's gradient produces the force field as shown above, an effective potential term is added such that its gradient produces a fictitious centrifugal force.
Resulting orbits can be all kinds of crazy shapes, but since they are not solutions of a simple 1/r potential field, they won't be Keplerian, and so you can't try to represent them with a central mass M.