As observed in a (now deleted) previous comment, the exponent of $\|\alpha\|_2$ should be $2k$ instead of $2$ for homogeneity reasons.

If the $\varepsilon_i$ are symmetric, then this can be proven by a variant of the exponential moment generating function method used to prove Khintchine's inequality. Indeed, if we normalise ${\bf E} \varepsilon_i^{2k}$ to be 1, then from Holder's inequality we see that ${\bf E} \varepsilon_i^j$ vanishes for odd $j$ and is bounded by $1$ for even $j$ up to $2k$. In particular, the exponential moment generating function

$$ {\bf E} \exp( t \varepsilon_i ) = \sum_{j=0}^\infty \frac{t^j}{j!} {\bf E} \varepsilon_i^j$$

is dominated by $\cosh( t^2 )$ in the sense that the coefficients of the former power series up to $t^{2k}$ are bounded in magnitude by those of the latter. $\cos(t^2)$ is dominated in turn by $\exp(t^2/2)$. Since

$$ {\bf E} \exp( t \varepsilon ) = \prod_{i=1}^n {\bf E} \exp(\alpha_i t \varepsilon_i )$$

we conclude that ${\bf E} \exp( t \varepsilon )$ is dominated by $\exp( \|\alpha\|_2^2 t^2 / 2)$. Extracting the $t^{2k}$ coefficient gives the claim.

The situation seems to be more subtle in the non-symmetric case; there does not seem to be a similarly simple argument (though one can certainly obtain a bound with $(2k-1)!!$ replaced by some weaker constant $C_k$ depending on $k$). It might be that the authors overlooked or neglected to mention a symmetry hypothesis when using this result.