$X^*(j\omega)$ is the complex conjugate of $X(j\omega)$. So if
$$X(j\omega)=X_R(\omega)+jX_I(\omega)$$
then
$$X^*(j\omega)=X_R(\omega)-jX_I(\omega)$$
and
$$X^*(-j\omega)=X_R(-\omega)-jX_I(-\omega)$$
where $X_R(\omega)$ and $X_I(\omega)$ refer to the real and imaginary parts of $X(j\omega)$, respectively. [Note that they are not the respective Fourier transforms of the real and imaginary parts of the time domain signal.]
Also, the notation $X^*(j\omega)$ is to be understood as $\left[X(j\omega)\right]^*$, i.e., the whole expression for $X(j\omega)$ is conjugated. Similarly for $X^*(-j\omega)$, i.e., $X^*(-j\omega)=\left[X(-j\omega)\right]^*$.
To gain some more understanding, consider the definition of the Fourier transform:
$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$$
So we have
$$X(-j\omega)=\int_{-\infty}^{\infty}x(t)e^{j\omega t}dt$$
and
$$X^*(-j\omega)=\int_{-\infty}^{\infty}x^*(t)e^{-j\omega t}dt=\mathcal{F}\{x^*(t)\}$$
as claimed in the quote of your question.