Let $X$ be a set and $A\subseteq X$. If I define the characteristic function of $A$ (relative to $X$) as the function $\chi_A: X\to \mathbb{F}_2$ such that
$$\chi_A(x) = \begin{cases}0; & x\not\in A \\ 1; & x\in A \end{cases}.$$
If $A, B\subseteq X$, then $\chi_A + \chi_B$ and $\chi_A\chi_B$ are both well defined functions from $X \to \mathbb{F}_2$
If $\chi_D = \chi_A + \chi_B$, then $\chi_D\left(x\right) = 1$ if exactly one of $\chi_A\left(x\right), \ \chi_B\left(x\right) = 1$, so $D = A\vartriangle B$.
I am trying to jump to the identity that:
$$C = A\cup B \iff \chi_C = \max\left(\chi_A, \chi_B\right).$$
Why is it necessary to show that $\mathbb{F}_2$ is an ordered field?
Simply because you couldn't have $\max$ in an unordered field.