Here is a more intuitive way to look at the electromagnetic field tensor: The electromagnetic field tensor is correlated to the Lorentz force law $$\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = e (\mathbf{E} + \mathbf{v} \times \mathbf{B}).$$ To problem with this equation is that it is not a "geometric equation", i.e. neither the righthand nor the lefthand side are frame independent.
Let $p^\alpha$ be the particles 4-momentum vector, the time shall be measured with the particles own clock. The frame-independent equation has the form $$\frac{\mathrm{d}p^\alpha}{\mathrm{d}\tau} = \frac{1}{\sqrt{1 - v^2}}e(\mathbf{E} + \mathbf{v} \times \mathbf{B}) = e(u^0 + \mathbf{u} \times \mathbf{B})$$$$\frac{\mathrm{d}p^\alpha}{\mathrm{d}\tau} = \frac{1}{\sqrt{1 - v^2}}e(\mathbf{E} + \mathbf{v} \times \mathbf{B}) = e(u^0\mathbf{E} + \mathbf{u} \times \mathbf{B})$$ The above equation is linear in the coordinates of the particle's 4-velocity $\mathbf{u}$, from where conclude that the lefthand side corresponds to a tensor quantity $\mathbf{F}$, named electromagnetic field tensor or Faraday tensor. $\mathbf{F}$ eats the particle's velocity vector $\mathbf{u}$ and puts out the electromagnetic 4-force vector.
One might wonder why $\mathbf{F}$ just takes one argument since a two-form usually eats to vectors and puts out a scalar. The answer to this is quite simple, just plug in another arbitrary 1-form $\mathbf{e}$ into $\mathbf{F}$ so that the output becomes a scalar: $$\mathbf{F}(\mathbf{e},\mathbf{u}) = \langle \mathbf{e}, \mathbf{F}(\mathbf{u})\rangle = \text{scalar}.$$ The Faraday tensor demonstrates the interrelation between the electric and magnetic field; neither one is frame-independent by itself, but by merging them into a tensor makes them frame-independent.