I don't guarantee that this is free of mistakes.
To compute the class group, we need to factor the ideals $(p)$ for primes $2 \le p \le 31$.
Case: $p = 2$. We have $(2) = (\sqrt{2})^2$ and $\mathcal{O}_L/(\sqrt{2}) \cong \mathbb{F}_2[\alpha]/(\alpha^2 - 1)$, so $(\sqrt{2}) = (\sqrt{2}, \alpha - 1)^2$. Hence
$(\sqrt{2}) = A^2$
where $A = (\sqrt{2}, \alpha - 1)$ is prime of norm $2$ and has order dividing $2$ in the class group.
The remaining primes are odd, and when working $\bmod p$ for $p$ odd we have $\mathcal{O}_L/(p) \cong \mathbb{F}_p[\sqrt{2}, \sqrt{-13}]$, so things simplify a little.
Case: $p = 13$. We have $(13) = (\sqrt{-13})^2$ and $\mathcal{O}_L/(\sqrt{-13}) \cong \mathbb{F}_{13}[\sqrt{2}]$ (note that $\bmod \sqrt{-13}$ we have $2 \alpha = \sqrt{2}$), which is $\mathbb{F}_{13^2}$. Hence $(\sqrt{-13})$ is prime and trivial in the class group.
The remaining primes do not ramify.
Case: $p = 7, 17, 31$. These are the primes for which $2, -13, -26$ are all quadratic residues. We can write them all up to sign as $a^2 - 2b^2$; consequently, $(p) = (a + b \sqrt{2})(a - b \sqrt{2})$, and $\mathcal{O}_L/(a + b \sqrt{2}) \cong \mathbb{F}_p[\sqrt{-13}] \cong \mathbb{F}_p^2$, so
$(a + b \sqrt{2}) = B_{1, p} B_{2, p}$
where the $B_{i, p}$ are distinct primes of norm $p$ and inverses in the class group. Similarly, $(a - \sqrt{2} b) = B_{3, p} B_{4, p}$.
Case: $p = 23$. We have $(23) = (3 + 4 \sqrt{2})(3 - 4 \sqrt{2})$ and $\mathcal{O}_L/(3 + 4 \sqrt{2}) \cong \mathbb{F}_{23}[\sqrt{-13}] \cong \mathbb{F}_{23^2}$. Hence the ideals $(3 \pm 4 \sqrt{2})$ are prime and trivial in the class group.
Case: $p = 29$. We have $(29) = (4 + \sqrt{-13})(4 - \sqrt{-13})$ and $\mathcal{O}_L/(4 + \sqrt{-13}) \cong \mathbb{F}_{29}[\sqrt{2}] \cong \mathbb{F}_{29^2}$. Hence the ideals $(4 \pm \sqrt{-13})$ are prime and trivial in the class group.
Case: $p = 3, 5, 11, 19$. These are the remaining primes for which $2$ is not a quadratic residue and exactly one of $-13, -26$ is. In this case $\mathbb{F}_p[\sqrt{2}, \sqrt{-13}] \cong \mathbb{F}_{p^2} \times \mathbb{F}_{p^2}$ and
$(p) = C_{1, p} C_{2, p}$
where the $C_{i, p}$ are distinct prime of norm $p^2$.
In summary, the class group is generated by the following prime ideals:
- $A$ (norm $2$, order dividing $2$),
- $B_{1, 7}, B_{3, 7}, B_{1, 17}, B_{3, 17}, B_{1, 31}, B_{3, 31}$ (norms $7, 7, 17, 17, 31, 31$),
- $C_{1, 3}, C_{1, 5}, C_{1, 11}, C_{1, 19}$ (norms $3^2, 5^2, 11^2, 19^2$).
To find relations between these elements we will compute norms of elements of $\mathbb{Z}[\sqrt{2}, \alpha]$:
- $N(\sqrt{2} + \alpha) = 11^2$, hence $C_{1, 11}$ is trivial.
- $N(\sqrt{2} + \sqrt{-13}) = 3^2 \cdot 5^2$, hence $C_{1, 3} = C_{1, 5}^{\pm}$.
- $N(3 + \sqrt{-26}) = 5^2 \cdot 7^2$, hence $C_{1, 5}$ is some product of the $B_{i, 7}$s.
- $N(14 + \sqrt{-13}) = 11^2 \cdot 19^2$, hence $C_{1, 19}$ is trivial.
- $N(1 + \alpha) = 2 \cdot 31$, hence $B_{1, 31} = A$. Conjugating, we find that $B_{3, 31} = A$.
- $N(3 + \alpha) = 2 \cdot 7 \cdot 17$, hence $A$ is some product of a $B_{i, 7}$ and a $B_{j, 17}$. Conjugating, we obtain another such product.
- $N(8 + \alpha) = 17^3$. We compute that $8 + \alpha$ is not divisible by $1 \pm 3 \sqrt{2}$, hence $B_{i, 17}$ has order dividing $3$. Together with the above, we conclude that $A = B_{i, 7}^3$.
- $N(\alpha) = 7^2$. We will use this below.
So the class group is generated by $B_{1, 7}$ and $B_{3, 7}$, both of which cube to $A$, which has order $2$. Now, the prime factorization of $(\alpha)$ is either (WLOG) $B_{1, 7}^2$ or $B_{1, 7} B_{3, 7}$. In the first case, $A = B_{i, 7}$ in the class group, so $A B_{i, 7}$ is trivial in the class group. It follows that some element of $\mathcal{O}_K$ has norm $14$, but using the tower property of the norm, the norm of any element of $\mathcal{O}_K$ has the form $a^2 + 26b^2$ ($a, b \in \mathbb{Z}$), and $14$ is not of this form.
Hence $(\alpha) = B_{1, 7} B_{3, 7}$, from which it follows that the class group is generated by $B_{1, 7}$, which has order $6$. Again using the tower property, there are no elements of norm $2$, so $A$ is nontrivial and $B_{1, 7}$ has order either $2$ or $6$. We ruled out the first possibility above, so:
The class group is cyclic of order $6$. It is generated by any prime lying over $7$.