Let's focus on the volatility contract price. Generalisation to cubic and quartic contracts is straightforward.
Following the paper's notations, the evaluation date is $t$ and the (European) contracts all expire at $T = t+\tau$. A volatility contract is specifically associated to the payoff function
$$ H[S] = R(t,\tau;S)^2 = \left(\ln S(t+\tau) - \ln S(t) \right)^2 = \left[ \ln \left(\frac{S}{S(t)}\right) \right]^2 $$
where we've used the paper's notation
$$ S(t+\tau) := S $$
According to arbitrage-free pricing theory, the price of a volatility contract should be calculated as
$$ V(t,\tau) = \mathcal{E}_t^*\left\{ e^{-r \tau} H[S] \right\} $$
where $\mathcal{E}^*_t\{ \cdot \}$ figures an expectation taken under the (risk-neutral) measure $\Bbb{Q}$ associated to the risk-free money market account numéraire, conditional on the information available at $t$.
The key result to conclude is the Carr-Madan formula (see references mentioned in the paper or here), which tells you that - for a sufficiently regular payout function - one can write out the conditional expectation above as
$$ \mathcal{E}_t^*\left\{ e^{-r \tau} H[S] \right\} = H[\bar{S}] + (S - \bar{S}) H_S[\bar{S}] + \int_{\bar{S}}^\infty H_{SS}[K] C(t,\tau;K) dK + \int_{0}^\bar{S} H_{SS}[K] P(t,\tau;K) dK \tag{3} $$
for any $\bar{S}$.
From the definition of $H[S]$, by differentiating we get
\begin{align}
H_S[S] &= 2 R(t,\tau;S) \frac{1}{S} \\
H_{SS}[S] &= \frac{2}{S^2}(1-R(t,\tau;S))
\end{align}
Now let's further simplify equation $(3)$ by picking $\bar{S} = S(t)$. This is a convenient choice since it means that terms involving $H[\bar{S}]$ and $H_S[\bar{S}]$ will disappear (because $R(t,\tau;\bar{S})=0$). We are then left with
\begin{align}
\mathcal{E}_t^*\left\{ e^{-r \tau} H[S] \right\} &= \int_{S(t)}^\infty \frac{2 \left(1-\ln\left(\frac{K}{S(t)}\right)\right)}{K^2} C(t,\tau;K) dK + \int_{0}^{S(t)} \frac{2\left(1-\ln\left(\frac{K}{S(t)}\right)\right)}{K^2} P(t,\tau;K) dK \\
&= \int_{S(t)}^\infty \frac{2 \left(1-\ln\left(\frac{K}{S(t)}\right)\right)}{K^2} C(t,\tau;K) dK + \int_{0}^{S(t)} \frac{2\left(1+\ln\left(\frac{S(t)}{K}\right)\right)}{K^2} P(t,\tau;K) dK \tag{7} \\
&= V(t,\tau)
\end{align}
That's it!
