To understand the equation in question consider how frequency offset $f_\mathrm{off}$ affects the received signal $r(t)$. For simplicity the authors assume the noise-free case and a frequency flat channel, i.e. the received signal before frequency offset is equal to the transmitted signal $s(t)$. Frequency offset is a frequency shift and can be expressed in time domain by
$$
r(t) = s(t) \mathrm{e}^{j2\pi f_\mathrm{off}t}.
$$
The received signal is now sampled at $t = n/F_\mathrm{s},\, n\in\mathbb{Z}$, where $F_\mathrm{s}$ is the sampling frequency. This yields
$$
r(n/F_\mathrm{s}) = s(n/F_\mathrm{s}) \mathrm{e}^{j2\pi n f_\mathrm{off}/F_\mathrm{s}}.
$$
In the following we write $r_n$ and $s_n$ for $r(n/F_\mathrm{s})$ and $s(n/F_\mathrm{s})$, respectively. The authors introduce the normalized frequency offset $\beta = f_\mathrm{off}/\Delta f$ but they don't define what $\Delta f$ is. From the context I understand that it is the subcarrier spacing $\Delta f = F_\mathrm{s}/N$, where $N$ is the number of subcarriers. Using this relationship we can rewrite the above equation as a function of $\beta$:
$$
r_n = s_n \exp\left(j2\pi \frac{\beta}{N}n \right).
$$
Next we consider a received cyclic prefix sample at $n$ (see above) and the corresponding received payload data sample $N$ samples apart at $n+N$:
$$
r_{n+N} = s_{n+N} \exp\left(j2\pi \frac{\beta}{N}(n+N) \right)
= s_{n} \exp\left(j2\pi\beta + j2\pi \frac{\beta}{N}n \right)
$$
In the above equation we have exploited that the cyclic extension is just a copy of the last $N_\mathrm{g}$ samples of an OFDM symbol and therefore $s_{n+N} = s_{n}$ (for all $s_{n}$ that are in the cyclic prefix). We can now see that the phase difference between the cyclic prefix and its corresponding data samples is $\arg(r_n^* r_{n+N})=2\pi\beta$ which is the answer to your first question. It is now easy to obtain an estimate $\hat\beta$ of $\beta$
$$
\hat\beta = \frac{1}{2\pi}\arg(r_n^* r_{n+N})
$$
It's just an estimation because in reality the received signal is corrupted by noise and channel distortion. The estimation can be improved by not only comparing one cyclic prefix and data sample but all $N_\mathrm{g}$ cyclic prefix samples with their respective copies. This is done by adding all $r_n^* r_{n+N}$ in the cyclic prefix and answers your second question about how the formula for $\beta$ is derived.
I have to say that the paper you're referring to is poorly written and that I probably wouldn't have understood it neither if I hadn't had some prior knowledge about this topic. Thus I suggest you also read the original work about this topic van de Beek et al.:
J. J. van de Beek, M. Sandell, and P. O. Borjesson, “ML estimation of time and frequency offset in OFDM systems,” IEEE Trans. Signal Process., vol. 45, no. 7, pp. 1800–1805, Jul. 1997.
