The particular signs in the transformation come from the realization of the Lorentz group on the field components ##\vec{E}## and ##\vec{B}##. You can understand this in two ways. The first is using the tensor representation of the electromagnetic field, i.e., the Faraday tensor in Minkowski space. In terms of the potentials its defined by
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
In 3D vector calculus on the other hand you have (I'm using Heaviside Lorentz units)
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A} - \vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
Now you can relate the ##F_{\mu \nu}## with ##\vec{E}## and ##\vec{B}## by splitting the Faraday tensor in temporal and spatial components (everything is of course in an inertial reference frame):
$$F_{0j}=\frac{1}{c} \partial_t A_j - \partial_j A_0=-\frac{1}{c} \partial_t A^j - \partial_j A^0=E^j,\\
F_{jk}=\partial_j A_k - \partial_k A_j=-\partial_j A^k + \partial_k A^j=-\epsilon^{jkl} B^l.$$
If ##{\Lambda^{\mu}}_{\nu}## is the matrix of the Lorentz transformation ##x'=\hat{\Lambda} x##, you can get the transformation of the ##\vec{E}## and ##\vec{B}## by applying the Lorentz transformation to the Faraday-tensor components and translate the result back into the 3D formalism:
$$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(x) = {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}( \hat{\Lambda}^{-1}x').$$
Another, much more elegant way is to introduce the complex valued 3D Riemann-Silberstein vector
$$\vec{F}=\vec{E}+\mathrm{i} \vec{B}.$$
You can show by direct comparison with the above given covariant method that this vector transforms under Lorentz transformations with a ##\mathrm{SO}(3,\mathbb{C})## matrix ##\hat{L}##. For real rotation angles you get of course a usual real orthogonal ##\mathrm{SO}(3)## matrix, and ##\vec{E}## and ##\vec{B}## of course transform as vectors under rotations. For purely imaginary rotation angles you get Lorentz boosts, and the imaginary rotation angle becomes ##\mathrm{i} \eta##, where ##\eta## is the rapidity of the boost, related to the velocity of the boost by ##\tanh \eta=v/c##, i.e.,
$$\vec{F}'(x')=\hat{L} \vec{F}(x)=\hat{L} \vec{F}(\Lambda^{-1} x').$$
