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