The definition for non-dividing is taken as the negation of the definition for dividing (as found in http://www.math.cmu.edu/~rami/simple.pdf : Definition 1.1 for example). Thus assuming $\varphi(x,c)$ doesn't divide over $A$, we have that either:
1) There is no infinite indiscrenible sequence that realizes the same type as the type of $c$ over $A$.
Or,
2) For every infinite indiscrenible sequence $I=(a_{i})$ that realizes the same type as the type of $c$ over $A$ and $\bigwedge\varphi(x,a_{i})$ is consistent.
Now, unless I have made a mistake, condition 1) can be considered as saying the type of $c$ over $A$ is algebraic. But what about conditon 2)? Is there a nice way thinking about this condition?
I looked at examples (For example $x<y$ in DLO doesn't divide over $\emptyset$, or $x\neq{x}$ in any theory). But I was wondering if there is a natural example that shows what exactly it is that 2) is meant to capture.
Edit 1: As pointed out in the comments, 1) implies 2). The question that I have here, rephrased would be, how should I be thinking about $\varphi(x,c)$ doesn't divide over $A$, provided that $c$ is not algebraic over $A$.
The notions of forking and dividing are meant to designate some formulas as "very small" (those which fork), in order to obtain a good notion of "generic behavior" over some parameters (the non-forking types: those which don't fall into any "very small" definable sets).
The condition that $\varphi(x,b)$ does not divide says that whenever we take a bunch of copies of the definable set, the $\varphi(x,b_i)$ where $(b_i)_{i\in \omega}$ is an indiscernible sequence, the set picked out by $\bigwedge_{i\in \omega}\varphi(x,b_i)$ is nonempty. That is, the definable set $\varphi(x,b)$ is "large enough" that its "translations" by automorphisms have a lot of overlap.
On the other hand, if $\varphi(x,b)$ divides, then we have a bunch of copies $\varphi(x,b_i)$, and there's some $k$ so that no $a$ can fall into more than $k$ of these copies. If we stretch the indiscernible sequence to be very long, say $\{b_i\}_{i\in I}$ then we have a huge number of distinct possible extensions of a type over $A$ to a type over $A\cup\{b_i\}_{i\in I}$ (the type "divides" into many many extensions) and it's reasonable to think of the formula $\varphi(x,b)$ that has so many incompatible copies as very small.
Now most reasonable notions of "smallness" describe ideals (the small sets are closed under subset and finite unions). The move from dividing to forking just takes the ideal generated by the dividing formulas (a definable set forks if and only if it's a subset of a finite union of sets which divide).
If you like measure-theoretic intuition, here's a way of formalizing the idea above that "dividing formulas are small". Let's suppose we have a finitely additive $[0,1]$-valued measure on the definable subsets of the monster model (this is called a Keisler measure), and further that the measure is automorphism invariant over $A$, i.e. if $b\equiv_A b'$, then $\mu(\varphi(x,b)) = \mu(\varphi(x,b'))$. We'd like to show that if $\varphi(x,b)$ divides over $A$, then $\mu(\varphi(x,b)) = 0$. Let the indiscernible sequence $(b_i)_{i\in \omega}$ witness dividing, and assume for contradiction that $\mu(\varphi(x,b))>0$. Since some conjunction $\bigwedge_{i = 1}^k \varphi(x,b_i)$ is inconsistent, there is some maximal $m$ such that $\mu(\bigwedge_{i = 1}^m\varphi(x,b_i)) > 0$. Then the countably many copies $\bigwedge_{i = 1}^m \varphi(x,b_{nm + i})$ for $n\in\omega$ all have the same positive measure, but their pairwise intersections have measure $0$. This contradicts the fact that the measure of the whole space is bounded above by $1$.