There is *no* difference between the formulas. The formula in the cyclotron radiation article is just the Larmor formula applied to the specific case of nonrelativistic charged particles moving in circles in a magnetic field.

For such particles, $F=ma$ combined with the Lorentz force law $F=qvB$ gives the acceleration as

$$a=\frac{qBv}{m}$$

so the [Larmor formula](https://en.wikipedia.org/wiki/Larmor_formula)

$$P=\frac{q^2a^2}{6\pi\epsilon_0c^3}$$

gives the radiated power as

$$P=\frac{q^2}{6\pi\epsilon_0c^3}\left(\frac{qBv}{m}\right)^2=\frac{q^4B^2v^2}{6\pi\epsilon_0m^2c^3}.$$

To see that this is the same as the [cyclotron radiation](https://en.wikipedia.org/wiki/Cyclotron_radiation) formula,

$$-\frac{dE}{dt}=\frac{\sigma_tB^2v^2}{c\mu_0},$$

just substitute the formula for the [Thomson cross section](https://en.wikipedia.org/wiki/Thomson_scattering),

$$\sigma_t=\frac{8\pi}{3}\left(\frac{q^2}{4\pi\epsilon_0mc^2}\right)^2$$

and the formula for the [vacuum permeability](https://en.wikipedia.org/wiki/Vacuum_permeability),

$$\mu_0=\frac{1}{\epsilon_0c^2}$$

to get

$$-\frac{dE}{dt}=\frac{8\pi}{3}\left(\frac{q^2}{4\pi\epsilon_0mc^2}\right)^2\frac{B^2v^2}{c}\epsilon_0c^2=\frac{q^4B^2v^2}{6\pi\epsilon_0m^2c^3}.$$

The total power $P$ given by the Larmor formula is the power radiated in all directions. The formula for the power per solid angle radiated in a specific direction is

$$\frac{dP}{d\Omega}=\frac{q^2a^2\sin^2\theta}{16\pi^2\epsilon_0c^3}.$$

Here $\theta$ is the angle between the acceleration vector and the direction in which the power is measured.

If you integrate this over all solid angles using $d\Omega=\sin\theta\,d\theta\,d\phi$, you'll get the Larmor formula for $P$ because

$$\sin^2\theta\,d\Omega=2\pi\int_0^\pi \sin^3\theta\,d\theta=2\pi\int_{-1}^1 (1-u^2)\,du=\frac{8\pi}{3}.$$