Definition. A type $p(x)$ for the theory $T$ is a collection of formulas with a free variable $x$ consitent with $T$.
-- see Michael Weiss' explanation of types.
Definition. A type $p(x)$ is principal if there is a formula $\alpha(x)$ such that $p(x)=\{ \psi(x) \mid \alpha \vdash \psi \}$
Definition. A type $p(x)$ is full if for every formula $\psi(x)$ either $\psi \in p(x)$ or $\neg \psi \in p(x)$.
A full type can be seen as a way of describing an a generic "element" of a theory. For any model $M$ and an element $m \in M$ we can consider the full type $p(m)$ generated by $m$. It contains all the formlas $\psi(m)$ that are true about $m$.
Omitting types theorem says that given a complete countable theory $T$ in a countable language, there is a model $M$ of $T$ which omits countably-many non-principal full types.
Why is the non-principal condition needed?
If a theory $T$ is complete, then any model of $T$ realizes all the full principal types. For if $p(x)$ is a complete principal isolated by $\psi(x)$, then $T \not\vdash \not \exists x.\psi(x)$, for otherwise the type $p(x)$ would be inconsistent with $T$. Because $T$ is complete it must be the case that $T \vdash \exists x. \psi(x)$. Then such $x$ realizes $p(x)$.
What happens if the theory $T$ is not complete?
Then it is not the case that all models of $T$ realize all the full principle types. For instance take Peano arithmetic, and take $T_1 = PA + Con(PA)$ and $T_2 = PA + \neg Con(PA)$. Both $T_1$ and $T_2$ are consistent, thus they have models $M_1$ and $M_2$. Then denote by $q(x)$ the type generated by the element $0$ in $M_1$, and by $p(x)$ the type generated by the element $0$ in $M_2$. Both of those types are isolated by the formula $\phi(x) := \neg \exists y. y < x$, because $\phi$ completely determines $0$. Albeit those types are types for different theories, they are both types for a common theory $PA$. Furthemore, those types are complete and isolated. However, one of the types states the consistency of $PA$ and the other one points out the inconsistency of $PA$. Thus both $q$ and $p$ cannot be realized in the same model.
