No, your first equation is not a consequence of your second one, nor vice-versa. The third one combines the first two.
The antisymmetric permutation symbols in n (spacetime) dimensions obey
$$ \epsilon^{j_1 \dots j_n}\epsilon_{i_1 \dots i_n} = n! \delta_{[ i_1}^{j_1} \dots \delta_{i_n ]}^{j_n} ,$$
where [...] denotes complete antisymmetrization of the lower indices.
As a result, contracting n-2 pairs, that is all indices but the leading two pairs, yields
$$ \epsilon^{\mu\nu j_3 \dots j_n}\epsilon_{\alpha\beta i_3 \dots i_n} = -(n-2)! (\delta_{ \alpha}^{\mu} \delta_{ \beta}^{\nu}- \delta_{ \beta}^{\mu} \delta_{ \alpha}^{\nu} ), $$
antisymmetric in all up and down free index pairs.
Your n=4 has a coefficient -2, while n=3 and n=2 a coefficient -1, and n=5 a coefficient -6, etc. This is all pure combinatorics, logically independent of the Lorentz group.
Now, the Lorentz group has an n(n-1)/2 -dimensional Lie algebra, (so 6d for n=4) indexed by the corresponding antisymmetric pairs of indices $\mu\nu$, etc,
$$
[J^{\mu\nu},J^{\rho \sigma}]=i(J^{\mu\rho} g^{\nu\sigma} +J^{\nu\sigma} g^{\mu\rho}- J^{\nu\rho} g^{\mu\sigma}-J^{\mu\sigma} g^{\nu\rho}),
$$
a unique form bearing all single and pairwise antisymmetries of the (pairwise antisymmetric) commutator on the left.
It is represented by all types of m × m matrices. in general, whose indices are suppresed here. But when it is represented by a "fundamental" set of n × n matrices, 4 × 4 ones in your case, with indices $\alpha\beta$ (antisymmetric, but symmetrized by the action of the metric if one of the two is 0--see below) it is straightforward, albeit tedious brute force,
to check that your first and third equations in fact satisfy it. I'd like to speculate that the 3rd eqn form is easier than the first, but I'd be lying... it's subjective.
In any case, this is only one of several bases in that dimension used, with the SO(3) subroup of rotations represented by sparse antisymmetric matrices and the three boosts in the coset by sparse symmetric ones.
In the mixed rotation/boost basis much of the compact structure you are observing is gone.
For n=3, your first formula still works for the 3 × 3 generators, but the third one is supplanted by
$$(J^{\mu\nu})_{\alpha\beta}=- i \epsilon^{\mu\nu\rho }\epsilon_{\alpha\beta\rho },\\
(M^\lambda)_{\alpha\beta}\equiv g^{\lambda\kappa}\epsilon_{\kappa\mu\nu} (J^{\mu\nu})_{\alpha\beta}=-2ig^{\lambda\kappa}\epsilon_{\kappa\alpha\beta} ,$$
one rotation (antisymmetric matrix $(M^0)^\alpha_{~~\beta}$); and two boosts ($(M^1)^\alpha_{~~\beta}, ~ (M^2)^\alpha_{~~\beta}$, now symmetric matrices, on account of the uneven action of the metric in raising the last index!).
*Intuition? I'm nonlinear at that... If you think of SO(n) instead of its noncompact brother the homogeneous Lorentz group SO(1,n-1), then the metric is the identity, and co/contra technically equivalent, so all is antisymmetric matrices and your 3rd equation is arguably more direct. It is yet another reformulation for your toolbox... Does it expedite Pauli-Lubanski for you?
