Confusion about the mathematical nature of Elecromagnetic tensor end the E, B fields I have quite a lot of confusion so the question may result not totally clear cause of that. I'll take any advice to improve it and I'll try to be as clear as possible. Everything from now on is what I have understood, so feel free to point out incorrectnesses.
I always considered the electric and magnetic fields $\bf E$ and $\bf B$ as vector fields, with this I mean that for example $E$ is an infinte set of arrows displaced one in each point of space. We represent that as 3-ple of numbers in each point of space.
Then I started to call vectors and vector fields different quantities, quantities that transform in this way under a Lorentz transformation: $A'^\mu = \Lambda^\mu \,_\nu A^\nu$, and I came to discover that neither $\bf E$ nor $\bf B$ transform like the spacial part of a 4-vector. So I wandered, what are they and how do they transform? They are the components of a different objects, the electromagnetic tensor here represented in his matrix form with metric (+---):
$$
F^{\mu \nu} \, = \,\left(\begin{matrix}0 & -E_x & -E_{y} & -E_{z} \\ E_{x} & 0 & -B_{z} & B_{y} \\ E_y & B_{z} & 0 & -B_x \\ E_z & -B_{y} & -B_{x} & 0\end{matrix}\right)\tag{0}
$$ 
So now I see: the components of 3-ple I used to consider are more generally the components of a tensor and transform like the components of a tensor.
Now I can write down the Lorentz force in two different forms:
$$
{\bf F}_L = q ({\bf E} + \frac{{\bf v}}{c} \times {\bf B})
\tag{1}$$
$$
 K^\mu = \frac{q}{c} F^{\mu \nu} u_{\nu}
\tag{2}$$
The spacial part of the latter equation is
$$
\frac{d \bf p}{ds} = \frac{q}{c} ({\bf E} u^0 + {\bf u} \times {\bf B})
\tag{3}$$
Now on the left hand side I have the spacial part of a 4-vector, while on the right hand side I have those ${\bf E}$ and ${\bf B}$ which aren't spacial parts of 4 vectors. But there are operations between these objects. For example there is a vector product I can write like 
$$
{\bf u} \times {\bf B} = \mathcal{\epsilon}_{ijk} u_j B_j
\tag{4}$$
Where I have written the components of the covector $u_\mu$ which numerically differs from those of the vector $u^\mu$ just by a minus sign. But has B that property even if it isn't the spacial part of a 4 vector? I mean lowering or raising its index changes its sign?
Considering the $0$-th component of the equation $(2)$ we encounter a scalar product 
$$
{\bf E} \cdot {\bf u} = E_x u^1 + E_y u^2 +E_z u^3
\tag{5}$$
What am I doing here? Am I just multiplying two 3-ples of numbers? I could do the same with the spacial part of the covector $u_\mu$
I notice it's unclear what I'm asking, the questions I need an answer to can be 
What are $E$ and $B$ mathematically and what is the nature of the mathematical operations involved? 
What am I doing in $(4)$ and $(5)$ just multiplying and summing numbers?
It's more like a request for a formal mathematical apparatus that relates vectors with n-ples, and $4$-vectors in Minkowski space with what I used to call vectors in $\mathbb{R}^3$?
Any help both in improving the question and in giving some insights are really appreciated.
 A: Great question! I had a lot of the same confusions when I first learned SR. 
First of all, $\vec{E}$ and $\vec{B}$ do transform as ordinary 3-vectors under rotations and translations. Mathematically, $\vec{E}$ and $\vec{B}$ are arrangements of components of the antisymmetric tensor $F^{\mu\nu}$. The reason professors are still justified in telling first-timers in electrodynamics that they are "vectors" is because they, considered as sets of objects in their own right, transform like a vector. If you want more detail, you should look into the group theory behind tensors (a favorite of mine is Zee). I can give you a brief overview (I don't know how much group theory you know, so I'll assume none. Forgive me if otherwise): A group is a set of objects which satisfies certain properties. In physics, the sets of transformations which preserve equations of motion (or, more properly, actions) are groups. One way to define 3-dimensional vectors is objects that transform in a certain way in a representation (a way of representing the abstract group action with matrices transforming a vector space) of the group $SO(3)$, orientation-preserving rotations in three dimensions (to be complete, we should include other transformations that preserve Euclidean distance, but rotations are good enough for now). In group theory, when you study these different representations, you'll find that some of them are isomorphic, which is just a highbrow way to say effectively equivalent. For $SO(3)$, the transformation of an antisymmetric 2-tensor (in the context of E&M, the spatial part of $F^{\mu\nu}$) turns out to be equivalent to a vector ($SO(3)$ is contained in the Lorentz group, so it's relevant for considerations in special relativity). This allows you to identify the correct arrangement of the components $F^{ij}$ as a vector, none other than $\vec{B}$. As for 4-vectors, we can get away with calling their spatial components 3-vectors because they transform in the same way under rotations. Also, for spatial transformations, indices up/down doesn't matter, since the spatial metric has the same sign in all of its components.
As for the dot and cross operations, they're really just bad notation. It means what you said it means, multiplying and adding/subtracting in a certain way. Once again, these operations go back to $SO(3)$ and transforming nicely under rotations (the dot product transforms as a scalar, the cross as a vector). The problem with writing them in relativistic equations is that it breaks manifest Lorentz invariance. Namely, the equations are still Lorentz invariant (more precisely, covariant, but who cares), but it's hidden by the bad notation. You'd have to work out the transformations under boosts by hand to make sure that everything worked out properly. This is where Minkowski 4-vectors are nice, because when you write the same equations in terms of 4-vectors, all you have to do is make sure the Lorentz indices line up. So the 3 versus 4-vector dilemma is a matter of notation; 4-vectors make Lorentz invariance obvious, 3-vectors obscure it. 
I should add one more note: there is a very deep differential-geometric interpretation of all of this involving differential forms and coordinate-independence, but I think it isn't as relevant to your immediate question. The upside of it is that it makes no reference to transformation laws, so if you're unsatisfied with "a vector transforms like a vector" etc, it can give some clarity. (See Frankel)
Hope this helped!
A: Both $\vec E$ and $\vec B$ transform as ordinary 3-vectors under rotations. It is under boosts that they transform not like the spatial parts of vectors, but they are both 3-vectors if you do not consider special relativity, but only classical rotational symmetry.
Once you consider special relativity, $\vec E$ and $\vec B$ are not good quantities anymore, precisely because they do not behave well under Lorentz transformations. But together, as components of the 2-form $F$, which transforms as a proper (0,2)-tensor under Lorentz transformations, they are "nice" objects again.
Actually, as I explain in this answer of mine, that the magnetic field can be thought of as a vector is a happy accident of 3 dimensions, and it should more properly be thought of as a spatial 2-form itself in a general framework.
To reconcile the four-force expression of the Lorentz force with the "usual" Lorentz force, i.e. reconcile your eq. (1) with eq. (3), note that they clearly agree in the frame where the particle is at rest, i.e. $u = (c,0,0,0)$ and then show that applying a Lorentz transformation to all objects involved reproduces both equations.
Nothing beyond these relations is going on here - I think what is confusing you is that you are trying to do relativity with manifestly non-covariant objects, i.e. $\vec E, \vec B$. This is not surprising - the power of covariant and invariant objects is precisely that they do not lead to such messy equations as non-covariant objects do. You should not expect to assign particular meaning to a non-covariant equation such as eq. (5).
