Why is the Faraday Tensor derived from the Lorentz force? If we start from the Lorentz force, $$\textbf{F}=q\textbf{E} +q\textbf{v}\times\textbf{B}$$
and use the four velocity u$^{\mu}$ and the four momentum p$^{\nu}$, then we get to $$\frac{dp^{\mu}}{d\tau}=qF^{\mu}_{\nu}u^{\nu}$$ OR $$\frac{dp^{\mu}}{d\tau}=qF^{\mu\nu}u_{\nu}$$.
Two questions:
1) Which one is the Faraday tensor? $F^{\mu}_{\nu}$ or $F^{\mu\nu}$?
What's the difference between the two and what is the significance of the latter being antisymmetric?
2) What is the "physical meaning" of deriving the Faraday tensor from the Lorentz force? 
And why does it relate to the four velocity?
Is it that there must be a momentum change in all frames?
 A: The short answer to (1). $F^\mu{}_\nu$ and $F^{\mu\nu}$ are related by $F^{\mu\nu} = g^{\nu\rho}F^\mu{}_\rho$ where $g^{\mu\nu}$ is the metric ($g^{\mu\nu} = \operatorname{diag}(1, -1,-1,-1)$ in Minkowski spacetime). Since the metric is invertible, either of $F^\mu{}_\nu$ and $F^{\mu\nu}$ uniquely determines the other. You can pick whatever version you wants, say $F^{\mu\nu}$ and call it "the Faraday tensor with up indices". If someone else wants it with down indices or both up and down, they just act on it with the metric in the appropriate way.
The long answer to (1). However the most natural variant is $F_{\mu\nu}$ with down indices. Then the anti-symmetry says that $F_{\mu\nu}$ is a 2-form. Why should $F_{\mu\nu}$ be a 2-form? Well, the EM field has 6 components, but we can introduce the 4-potential, which has 4, and is a 4-vector. A 1-form is precisely a vector with down index; $A_\mu$ is a 1-form. The EM field should be the derivative of the potential in some sense. There is a differential operator called the exterior derivative, denoted $d$, that makes 2-forms out of 1-forms. It's sort of a generalized curl. Then we could guess that maybe $F_{\mu\nu} = (dA)_{\mu\nu}$, and this agrees with the usual definition. $d$ is a very neat operator, because $d(d\eta) = 0$, always. This is a generalization of curl grad = 0 and div curl = 0. In the present context that means that we automatically get two of Maxwell's equations. 
Maybe that's not terribly useful because it's more or less a game with index manipulation. But Elie Cartan showed that $d$ makes sense in any spacetime, even the curved ones of general relativity. If we think about Maxwell's equations in terms of the 1-form $A_\mu$ (or the 2-form $F_{\mu\nu}$) and the operator $d$, we automatically get a theory of electromagnetism that works even in curved spacetime.
There is a very high-level way of thinking about $F_{\mu\nu}$ where you think about it as a curvature in a specific sense. Now, the curvature in general terms is something that measures the failure of derivatives to commute. So you put in two vectors $x^\mu$ and $y^\mu$, and the curvature should tell you how much taking derivatives along $y^\mu$ first and then along $x^\mu$ differs from doing it in the other order. The curvature should take two vectors and switching their order should change the sign of what comes out. But this is the definition of a 2-form.
But why would one ever want to think about the electromagnetic field as a curvature? You can do electrodynamics in practice perfectly fine without ever making that interpretation, but it opens up a new range of tools you can use to think about Maxwell's equations. And, based on this curvature description, there is a vast generalization of Maxwell's equations called gauge theory, which makes up all of the standard model of particle physics. I recommend the book Gauge Fields, Knots and Gravity by Baez and Muniain. It's a fairly advanced book but you do not strictly need much more some familiarity than special relativity and Maxwell's equations to read it, if you put in enough work.
(2) I am not sure what you mean by deriving the Faraday tensor from the Lorentz force. Well, if you assume that the equation of motion has the form $$\frac{dp^\mu}{d\tau} = q F^\mu{}_\nu u^\nu$$
and choose a reference frame, then you can read off what the components of $F^{\mu}{}_\nu$ should be to match the Lorentz force.
Why is it the 4-velocity that's involved? It's similar to what you say, that "there must a momentum change in all frames". The laws of physics should be the same in all frames, and since we have observed the Lorentz force, the relativistic equation of motion must somehow involve the velocity, right? But the velocity on its own is not a proper relativistic quantity, we have to use the 4-velocity. Then I guess you can justify the assumption $$\frac{dp^\mu}{d\tau} = qF^\mu{}_\nu u^\nu$$ 
because certainly the Lorentz force is linear in $\mathbf v$.
A: In the relativistic theory, you suppose a metric tensor to be $g^{\mu\nu}=g_{\mu\nu}$ such that $\left(ds\right)^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$ is the infinitesimal length element. For special relativity, you define $g_{11}=g_{22}=g_{33}=-1$ and $g_{00}=1$. The $0$-component stands for the time and the $1,2,3$-components stand for the space (other conventions exist, this one may be the most popular). You then define a quadric-vector $x^{\mu}\equiv\left(ct,\boldsymbol{x}\right)=\left(ct,x,y,z\right)$ and of course $x_{\mu}=g_{\mu\nu}x^{\nu}\equiv\left(ct,-\boldsymbol{x}\right)$ Then you have 
$$F^{\mu\nu}=g^{\mu\lambda}F_{\lambda}^{\;\nu}$$
which answers your first question.
Next your second question. It is more customary to define the field-tensor $F^{\mu\nu}$ as 
$$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$
with $\partial_{\mu}=\partial/\partial x^{\mu}\equiv\left(\partial_{ct},\boldsymbol{\nabla}\right)$ and $A^{\mu}=\left(V/c,\boldsymbol{A}\right)$. Once again, you have $A_{\mu}=g_{\mu\nu}A^{\nu}$, ... so the space component change sign. This is nothing more than the usual definitions for the electric $\boldsymbol{E}$ and magnetic $\boldsymbol{B}$ field 
$$\boldsymbol{E}=-\dfrac{\partial\boldsymbol{A}}{c\partial t}-\boldsymbol{\nabla}V\;\;\text{and}\;\;\boldsymbol{B}=\boldsymbol{\nabla\times A}$$ 
in terms of the potentials $V$ and $\boldsymbol{A}$. Calculating it explicitly, you can write the field-tensor into a matrix
$$F^{\mu\nu}\equiv\left(\begin{array}{cccc}
0 & -E_{x} & -E_{y} & -E_{z}\\
E_{x} & 0 & -B_{z} & B_{y}\\
E_{y} & B_{z} & 0 & -B_{x}\\
E_{z} & -B_{y} & B_{x} & 0
\end{array}\right)$$
if you wish. From this you can verify that the Lorentz equation is verified in the relativistic form 
$$\dfrac{dp^{\mu}}{d\tau}=\dfrac{q}{c}F^{\mu\nu}v_{\nu}$$
provided you defined $p^{\mu}\equiv\left(E/c,\boldsymbol{p}\right)=m\left(v_{0},\boldsymbol{v}\right)$. 
More details can be found in 

Classical electrodynamics J.D. Jackson, chapter 11, 3-nd edition.

