Let $G=GL_2(F)$ where $F$ is a non-Archimedean local field of characteristic $0$, for example $\mathbb{Q_p}$. Let $\chi=1_T$ be the trivial character of the maximal split torus $T=\begin{pmatrix}* & 0 \\ 0 & * \end{pmatrix}$ of $G$. We can smoothly induct $\chi$ to a representation of $G$, we call this $\text{Ind}1_T$. By the irreducibility criterion (lemma 9.6 of "The Local Langlands Conjecture for GL(2)" by Bushnell and Henniart), the smooth representation $\text{Ind}1_T$ has $G$-length 2, and a one-dimensional $G$-subspace $1_G=1_T \circ \det.$ We call the irreducible $G$-quotient of $\text{Ind}1_T$ the Steinberg representation, and denote it by $St_G$. I am trying to see why, in this case, $(St_G)_N\simeq\delta_B^{-1}$ where $\delta_B$ is the module character of $B$, and $(St_G)_N$ the Jacquet module of $St_G$ at $N$.
I know that there is an exact sequence \begin{equation} 0 \rightarrow 1_G \rightarrow \text{Ind}1_T \rightarrow St_G \rightarrow 0 \end{equation} hence after applying the exact Jacquet functor, and noting $1_T=(1_G)_N$, we get \begin{equation} 0 \rightarrow 1_T \rightarrow (\text{Ind}1_T)_N \rightarrow (St_G)_N \rightarrow 0. \end{equation}
Also, by the induction-restriction lemma of the same book mentioned above, the Jacquet module of $\text{Ind}1_T$ fits into the following exact sequence: \begin{equation} 0 \rightarrow \delta_B^{-1} \rightarrow (\text{Ind}1_T)_N \rightarrow 1_T \rightarrow 0. \end{equation} I'm missing how to conclude from here that $(St_G)_N$ and $\delta_B^{-1}$ are isomorphic.
First, note that the Steinberg representation is more usually defined via normalized induction, so that it is tempered. However, the argument in that case is along exactly the same lines as in your case (as a heuristic, note that what you've written so far is at least true on the level of Grothendieck groups), and you have essentially all the pieces in your question. For a description of the normalized case, see for examples Ngo's notes on his website.
Now, to conclude from what you have written, it suffices to show that your second short exact sequence is split, which is a general fact. The map $$ \mathrm{Ind}(1_T)_N\to 1_T $$ is given by $f\mapsto f|_B$, where we think of $\mathrm{Ind}(1_T)_N$ as functions on $B\setminus G/N$. In this case a splitting is given simply by the constant functions. Thus $$ \mathrm{Ind}(1_T)_N\simeq \delta_B^{-1}\oplus 1_T. $$ As $\delta_B^{-1}\neq 1_T$, the claim follows.