Let $X$ be a smooth projective variety over some field $k$. Then each closed point $x$ has an associated residue field $k(x)$ which is a finite extension of $k$ and a point is rational when $k(x)=k$.
What happens with divisors? If $D$ is a divisor on $X$, with irreducible components $D_j$, is it possible to associate a finite extension of $k$ to each $D_j$?
What is the relation then between the fields of $D_i$, $D_j$ and $D_i \cap D_j$?
Here's an alternative construction that doesn't require moduli spaces. For simplicity, let's assume that the field $k$ is perfect, let $\bar k$ be an algebraic closure of $k$, and let $G_k=\mbox{Gal}(\bar k/k)$. For any subvariety $Y\subset X$ defined over $\bar k$, let $G(Y)=\{\sigma\in G_k : \sigma(Y)=Y\}$. Then the natural field to associate to $Y$ is $k_Y:=\bar k^{G(Y)}$, i.e., the fixed field of $G(Y)$. In particular, you can apply this construction to your $D_i$.
As for your last question, first you need to define $k_Y$ for subschemes that might not be irreducible or reduced, since your intersection $D_i\cap D_j$ is a union of subvarieties of codimension 2 (and it really should be treated as a subscheme, not a simple set theoretic intersection). But a similar Galois-theoretic construction should work, after which Will's inclusion $k_{D_i\cap D_j}\subseteq k_{D_i}k_{D_j}$ follows from basic Galois theory.
A final note. If, say, $D=D_1+D_2$, then $k_D$ may be strictly smaller than $k_{D_1}\cap k_{D_2}$. For example, taking divisors in $\mathbb{P}^1_{\mathbb Q}$, let $D_1=(i)$ and $D_2=(-i)$. Then $k_{D_1}= k_{D_2}=\mathbb{Q}(i)$, while $k_D=\mathbb{Q}$.
One can answer this sort of question using a moduli space of divisors. Since each divisor corresponds to a point on that moduli space, one just takes the residue field of that point. This is also equal to the smallest field over which equations for the divisor can be written down.
If this was asked about divisor classes, that moduli space would be the Picard scheme. To get a moduli space of divisors, we first construct a moduli space of effective divisors as a projective space bundle on the Picard scheme, then take the self-product, since each divisor can be written as a pair of effective divisors, then remove the closed subset where the same irreducible component appears with both positive and negative sign.
I think the only relation is that $k(D_i \cap D_j) \subset k(D_i)k(D_j)$, since from equations for $D_i$ and equations for $D_j$ one can write down the equations for $D_i \cap D_j$.
