Not every central force admits a circular orbit since only an attractive interaction can balance the repulsive centrifugal term. On the other hand, every attractive force has a circular orbit since by an appropriate choice of angular momentum the centrifugal term, $L^2/2\mu r^2$, can be chosen to cancel the attractive interaction, $U(r)$. Finally, attractive potential is not a sufficient for the existence of stable orbits.
The circular orbit of radius $r_0$ is stable if and only if $U_{\mathrm{eff}}(r_0)$ corresponds to a minimum of the effective potential,
$$U_{\mathrm{eff}}(r)=\frac{L^2}{2\mu r^2}+U(r).$$
This implies that the second derivative of $U_{\mathrm{eff}}$ at $r_0$ must be positive. Hence,
$$U''(r_0)>-\frac{3L^2}{\mu r_0^4}.\tag1$$
For circular orbits, the radial effective force (which includes interaction and centrifugal forces) vanishes and then $U_{\mathrm{eff}}(r_0)'=0$. Thus
$$r_0^3=\frac{L^2}{\mu U'(r_0)}.$$
Plugging this into Eqn (1) we obtain the following condition
$$\frac{U''(r_0)}{U'(r_0)}+\frac{3}{r_0}>0,$$
for stable orbits. As an example, assuming an attractive potential with a power law $U=kr^{n}$, the last equation gives us that is has stable orbits only for $n>-2$.
It is worth mentioning that there are different concepts of stability regarding orbital motion. The one assumed here, for which the above result holds, says that a circular orbit is stable if it remains bounded under small perturbations (a bounded orbit is one whose radius is limited by $r_{\mathrm{min}}\leq r\leq r_{\mathrm{max}}$). A different and also common concept is Lyapunov Stability. In this case, among all power law central forces, only the harmonic oscillator gives stable orbits.
You're forgetting that for fixed $m$, $u_0$ is a function of $L$. $J(u,L)$ is function of two variables, not one, so your differential equation for "$J'(u)$" isn't so simple.
If we take $m$ to be constant, then in principle $u$ and $L$ are independent parameters. The function $J(u,L)$ is defined via
$$J(u,L)=-\frac{m}{L^2u^2}f\left(\frac{1}{u}\right) \tag{1}$$
where $f(r)$ is some random force function. In the discussion you quoted, we are expanding about an arbitrary point $u_0$ defined by the following constraint equation:
$$u_0=J(u_0,L) \tag{2}$$
Here we can solve for $u_0$ in terms of $L$, to get $u_0(L)$, which we will assume to be single-valued for simplicity. This is intuitive: for a specified value of angular momentum $L$, the value $u_0(L)$ gives a circular orbit. Likewise, we can flip this around write:
$$u=J(u,L_0)\longrightarrow L_0(u)\tag{3}$$
For a specified radius $u$, the angular momentum $L_0(u)$ gives a circular orbit.
In that wikipedia article, they found that $\beta$ is independent of $L$ (or equivalently $u_0$, if $u_0(L)$ is a one-to-one function):
$$\frac{\partial J(u_0(L),L)}{\partial u}=1-\beta^2 \tag{4}$$
You cannot integrate this equation with respect to $u$ because there is no $u$ involved actually - we are evaluating at $u=u_0(L)$. It's a function of $L$. Likewise we can rewrite (4) as
$$\frac{\partial J(u,L_0(u))}{\partial u}=1-\beta^2 \tag{5}$$
Still, you cannot simply integrate and obtain $J(u,L_0(u))$ because that's not how the chain rule works:
$$\frac{d}{du}J(u,L_0(u))=\frac{\partial}{\partial u}J(u,L_0(u))+\frac{\partial J(u,L_0(u))}{\partial L} \frac{dL_0}{du} \tag{6}$$
So you still need to calculate $\frac{dL_0}{du}$ in order to integrate and obtain $J(u,L_0(u))$.
What the article does is use a clever manipulation to derive an equation that does away with explicit $L$ dependence, leaving a single-variable differential equation in $u$ to be solved.
$$\frac{\partial J (u_0(L),L)}{\partial u}=1-\beta^2=-2+\frac{u_0(L)}{f\left(\frac{1}{u_0(L)}\right)}\left[\frac{d}{du}f\left(\frac{1}{u}\right)\right]_{u=u_0(L)} \tag{7}$$
In (7), we find that all the dependence on $L$ is through the function $u_0(L)$. Since $u_0(L)$ is assumed to be a smooth function of $L$, we can really just drop the $L$ and solve for $u_0$. This is equivalent to having derived
$$\frac{\partial J (u,L_0(u))}{\partial u}=1-\beta^2=-2+\frac{u}{f\left(\frac{1}{u}\right)}\frac{d}{du}f\left(\frac{1}{u}\right)\tag{8}$$
This gives a differential equation for $f(r)$ which we can solve, and work back to get $J(u,L)$.
Best Answer
The orbit is in 2d and "oscillates" between a minimum and a maximum $r$. The position in the plane if given by $(r(t),\phi(t))$ but here $t$ has been eliminated and you have $r(\phi)$.
As you go once from $r_{min}$ to $r_{max}(\phi)$, the body will advance along the orbit by an angular distance $\Delta \phi$. As you go from $r_{min}$ to $r_{max}$ and back to $r_{min}$, you advance by an angle $2\Delta \phi$.
To get a closed orbit, you must eventually come back to your starting point, meaning you must make an integer number $b$ of trips between $r_{min}$ and $r_{max}$ while advancing by an integer multiple $a$ of $2\pi$. This is the geometrical origin of the $2\pi a/b$ factor.
Edit: In answer to a comment, two situations are illustrated below. In both cases $r_{min}=1$ and $r_{max}=3$, and these values are shows as red thick lines. These values restrict the orbits to a ring of inner radius $1$ and outer radius $3$. The radius oscillates between $1$ and $3$ with some frequency $\omega_r$, as can be seen by the black lines in the figures.
The parametric equations for the figures on the left and the right are, respectively, $$ r(\phi)=2+\cos\left(\sqrt{3}\phi\right)\, ,\qquad \hbox{and}\qquad r(\phi)=2+\cos\left(\phi\right) $$
In the first case, the ratio $\omega_\phi/\omega_r$ is not commensurate since $\sqrt{3}$ is irrational, and the orbit does not close. The best way to see this is to note that the start of the parametric curve is at $r=3,\phi=0$ but, at the end of the curve, $r\ne 3$. Because the ratio $\omega_\phi/\omega_r$ is irrational, the orbit would eventually densely fill the ring.
In the second case, on the other hand, the ratio $\omega_\phi/\omega_r$ is commensurate, and one can show (if we follow the curve through its $\phi$ evolution) that in fact it goes from $r_{min}\to r_{max}\to r_{min}$ exactly once when $\phi$ goes from $0\to 2\pi$.