Help with definition: partition mod ultrafilter. 
I do not see how the partition is well-defined.
By definition $A\neq\varnothing\mbox{ mod }D\iff A\notin I_{D}$.
Since D  is a maximal filter $A\notin I_{D}\iff A\in D$ .
So $A\neq\varnothing\mbox{ mod }D\iff A\in D$ .
Thus, the first bullet says $\mathcal{A}\subseteq D$ . So in particular if $A,A'\in\mathcal{A}$ , $A\cap A'\in D$ . But the second bullet says for distinct $A,A'$ , we have $A\cap A'\in I_{D}\implies[A\cap A']^{C}\in D$ , which is impossible.
EDIT: $\mathcal G$ is independent modulo $D$ iff for all $l\in\omega$, forall  {$g_1,...,g_l$}$\subseteq \mathcal G$ and {$j_1,...,j_l$}$\subseteq \omega$, we have {$n\in\omega:g_k(n)=j_k(n),1\leq k\leq l$}$\neq\varnothing$ mod $D$.

Kramer, Linus; Shelah, Saharon; Tent, Katrin; Thomas, Simon "Asymptotic cones of finitely presented groups".  Adv. Math. 193 (2005), no. 1, 142–173.

The paper can be found here - Go to section 4. The "partition" definition is after Lemma 4.4.
 A: As was guessed in some of the comments, $D$ need not be an ultrafilter. The maximality statement in the quoted passage is intended to mean that $D$ is maximal among those filters modulo which $\mathcal G$ is independent.  Unless $\mathcal G$ is empty, it cannot be independent modulo an ultrafilter.
To understand the general notion of partition modulo a filter (which makes sense for any filter, whether or not maximal with respect to some $\mathcal G$), it may be best to begin with the special case where $D$ is just $\{\omega\}$. Then the three bullet items say that (1) the partition consists of nonempty sets, (2) distinct sets in the partition are pairwise disjoint, and (3) the sets in the partition cover all of $\omega$ (that's easily equivalent to the literal reading: every nonempty subset of $\omega$ meets some set in the partition).  So this is just the familiar notion of a partition of $\omega$.  
When $D$ is not just $\{\omega\}$, one more ingredient goes into the definition: You are allowed to restrict attention to any set in $D$; equivalently, you are allowed to ignore anything that happens in a set in $I_D$.  So the first bullet item says that the sets in the partition (I'll call them "blocks" for short) are not merely nonempty but remain so even when intersected with any element of $D$.  So the effect of $D$ is to make this requirement more stringent than it was in the preceding paragraph.  The second bullet item says that any two distinct blocks, though perhaps not actually disjoint, become disjoint when intersected with a suitable set in $D$ (which may depend on the particular blocks in question).  So this second requirement has become less stringent.  Finally, the third bullet item says that all the blocks together, though they might not cover all of $\omega$, must cover a set in $D$; thus, by restricting attention to that set in $D$, we no longer see any failure of covering.  So the third requirement has also become less stringent.  Altogether, the definition says that, if you focus on "almost all" of $\omega$, in the sense of $D$, then $\mathcal A$ will look like a partition.
This informal description can be made formal by abstracting a bit to deal with Boolean algebras.  Specifically, the Boolean algebra $\mathcal P(\omega)$ of all subsets of $\omega$ has a quotient Boolean algebra $\mathcal P(\omega)/D$, obtained by identifying any two subsets of $\omega$ that agree almost everywhere with respect to $D$.  That is, identify $a$ with $b$ if there is $d\in D$ with $a\cap d=b\cap d$.  Then a partition $\mathcal A$ modulo $D$ is almost the same as a "partition of $1$" in this quotient algebra, i.e., a family of non-zero, pairwise disjoint elements of $\mathcal P(\omega)/D$ whose join (= supremum) is the top element $1$ of the algebra.  The "almost" in the preceding sentence refers to the fact that a partition if $1$ is a family of elements of the quotient algebra $\mathcal P(\omega)/D$, i.e., a family of equivalence classes, while $\mathcal A$ is a set of representatives for such a family, one element from each of the equivalence classes in the family.
