I think I understand what Kevin is saying in the Theorem. Let  $\mathfrak o$ be the ring of integers of $K$ and $\mathfrak p$ the maximal ideal. Define the following subgroups of 
$G=GL(2, K)$:


$I= \begin{pmatrix}\mathfrak o^{\times} & \mathfrak o //
    \mathfrak p & \mathfrak o \end{pmatrix}$ 

$I_n=\begin{pmatrix} \mathfrak o^{\times} & \mathfrak o // \mathfrak p^n & 1+\mathfrak p^n\end{pmatrix}$  


$Z_n= \begin{pmatrix}1+\mathfrak p^n & 0 // 0 & 1+\mathfrak p^n\end{pmatrix}$

Lemma. Let $\pi$ be a smooth representation of $I$, such that $Z_n$ acts trivially, $Z_{n-1}$ does not act trivially and 
the space of $I_n$-invariants is non-zero. Then the restriction of 
$\pi$ to $I_n Z_0$ contains a one dimensional subrepresentation of the 
form  $\chi: \begin{pmatrix} a & b // c & d\end{pmatrix} \mapsto \chi_1(d)$, where
$\chi_1: \mathfrak o^{\times}\rightarrow \mathbb C^{\times}$ is a smooth character of conductor $\mathfrak p^n$.

Proof. Look at the action of the abelian group $Z_0$ on $\pi^{I_n}$.

The pair $(I_nZ_0, \chi)$ is a type for the Bernstein component, which contains the principal series representations that Kevin describes. In other words, if $\pi$ is an 
irreducible smooth representation of $G$, then $Hom_{I_n Z_0}(\chi, \pi)\neq 0$ if and only if $\pi$ is a principal series rep with one character unramified, the other of conductor $\mathfrak p^n$. For this you could look either in the appendix by Heniart to:

http://www.math.u-psud.fr/~breuil/PUBLICATIONS/multiplicite.pdf

or in the book of Bushnell and Henniart. The main point being that the representation $\chi$ (as a representation of 
$\begin{pmatrix} \mathfrak o^{\times} & 0 // 0 & \mathfrak o^{\times}\end{pmatrix}$)
shows up in the $U$-coinvariants of $\pi$, where $U$ is unipotent upper (lower?) triangular matrices.