When the splitting fields of shifted generic polynomials are linearly disjoint? Let me start by rigorously pose my question. 
Let $K$ be an algebraically closed field of characteristic $2$, let $n$ be an even integer number, let $f(X) = X^n + T_1 X^{n-1} + \cdots + T_n$, be the generic polynomial, that is, $T = (T_1, \ldots, T_n)$ is a tuple of algebraically independent variables over $K$. 
Let $\Omega = $ { $\omega_1, \ldots, \omega_m$} be a finite subset of $K$, let $f_i(X) = f(X) - \omega_i$, and let $F_i$ be the splitting field of $f_i$ over $K(T)$ ($i=1,\ldots, m$).
Question: For which $\Omega$ the splitting fields $F_1, \ldots, F_m$ are linearly disjoint over $K(T)$?
Remarks:


*

*If the characteristic of $K$ is NOT $2$, or if $n$ is odd, then the splitting fields are linearly disjoint for arbitrary $\Omega$. Thus, I pose the question the specific case of $p=2$ and $n$ even.

*The answer cannot be ALWAYS, as in the previous remark. Indeed, one can show that if $p=n=2$, $m=4$, and $\omega_1 + \omega_2 + \omega_3 + \omega_4 = 0$, then the splitting fields are not linearly disjoint. In fact, if $p=n=2$, the answer is that the splitting fields are linearly disjoint if and only if the sum of any even number of elements of $\Omega$ does not vanish. 

*How one proves 1 + 2: The linear disjointness of the splitting fields can be reduced to the linear independent of the discriminant as elements in $H^1(K,\mathbb{Z}/2\mathbb{Z})$. If $p\nmid n$, then one can use ramification theory to achieve this, if $p\neq 2$ but divides $n$, one can calculate this by hand using the formula given by the determinant of the Sylvester matrix. If $p=2$, I know of no formula for the discriminant in terms of the coefficients. However when $p=n=2$ situation is simple enough to do calculations and hence get 2. 
Motivation: The linear disjointness of the splitting fields allows one to calculate a Galois group of a composite of polynomials, which in turn yields arithmetic features of the ring of polynomials over large finite fields. Let me not elaborate on that here 
 A: Lior, you say:

If $p=2$, I know of no formula for the
discriminant in terms of the
coefficients.

However, this link says that the Sylvester matrix works the same over every field, whether $p=2$ or not. The article on PlanetMath about determinants seems to support this, not including a requirement of $p\neq2$. Using the determinant Sylvester matrix you get a formula in terms of the polynomial's coefficients like you wanted; I don't see why counting a determinant over a field with $p=2$ should be any different than in any other field, of course other than the fact that your definition of '$+$' is different. It doesn't seem you need anything other than to know how to differentiate polynomials over your field.
On the other hand, the Wikipedia article on the determinant also says that the determinant of quadratic forms cannot be generalized for fields of characteristic $p=2$. Is this what you were thinking of?
I have a feeling an answer to your question might be in the book "Generic Polynomials. Constructive Aspects of the Inverse Galois Problem" By Jensen, Ledet, Yui.
You might also find inspiration in the paper "Computing Galois Groups of Polynomials through ODEs", by Cormier, Singer, Trager, Ulmer, 2000, wherein they construct a linear differential operator that takes a polynomial and spits out information about its Galois group over the field as well as its algebraic closure; you might be able to come up with some form of chain rule for composite polynomials, solving your original problem. The paper was submitted to Journal of Symbolic Computation.
