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Use unnormalized parabolic induction, the convention fixed by the supplied sequence and by the occurrence of constants in $V(1)$. Its functions satisfy $f(bg)=\chi (b)f(g)$, and $G$ acts by right translation. Characters in this smooth category are smooth characters. Put $\lambda ={\chi }^w{\delta }^{-1}$, retaining the question's modulus $\delta (\mathrm{diag}(a,d))=|d/a|$.

First derive the evaluation adjunction for unnormalized induction. A $G$-intertwiner $A:V(\chi )\to V(\xi )$ gives a functional $\ell (v)=A(v)(1)$. Since $A$ intertwines and induced functions have left covariance, $\ell (\pi (b)v)=\xi (b)\ell (v)$. Conversely such a functional definesThis function has the required left covariance. Any compact-open subgroup fixing $v$ fixes it on the right, so it is a smooth induced function; compactness of $B\backslash G$ removes any additional support issue. The construction is $G$-equivariant, and the two operations are inverse. Since $\xi$ is trivial on $U$, these functionals factor precisely through the Jacquet module, givingNow use the given exact sequence $0\to \lambda \to J\to \chi \to 0$, where $J=V(\chi {)}_U$ has dimension two.

If $\lambda \ne \chi$, choose $t_0\in T$ on which their values differ. Its operator on $J$ is upper triangular with distinct eigenvalues, so has two one-dimensional eigenspaces. Since $T$ is abelian, all its operators commute with this operator and preserve the eigenspaces. They are the linear character lines $\lambda$ and $\chi$. This proves that distinct-character extensions of an abelian group split, so $J\cong \lambda \oplus \chi$. A map between one-dimensional linear character spaces is zero unless the linear characters agree, and otherwise is one-dimensional. Therefore the required dimension is one for $\xi =\lambda$ or $\xi =\chi$, and zero for any other linear character.

The case $\lambda =\chi$ needs an extra argument: the exact sequence alone would also permit a split self-extension with two quotient functionals. Write $\chi (\mathrm{diag}(a,d))={\chi }_1(a){\chi }_2(d)$. Equality with $\lambda$ forces ${\chi }_1/{\chi }_2=|\cdot |$, so $\chi =(\eta \circ \det ){\chi }_0$, where $\eta ={\chi }_2$ and ${\chi }_0(\mathrm{diag}(a,d))=|a|$. Multiplication of induced functions by $\eta (\det g)$ identifies this with a determinant-character twist of $V({\chi }_0)$. Twisting is invertible, so it is enough to prove nonsplitting for ${\chi }_0$.

Use the open Bruhat decomposition coordinateFor $x\ne 0$ the matrix factorizationshows that $g_f(x)=|x{|}^{-1}f(1)$ for all sufficiently large $|x|$: the lower-unipotent matrix approaches the identity, where $f$ is locally constant. The coordinate map is injective, since this open cell is dense in the projective-line quotient and induced functions are locally constant sections.

For $t=\mathrm{diag}(a,d)$, direct multiplication givesThere exists a section $f_0$ with $f_0(1)=1$ and coordinate functionIndeed the displayed factorization changes to the lower chart near infinity, where this tail corresponds to the constant function one. Thus it extends smoothly across the missing point. With $t=\mathrm{diag}(\varpi ,1)$ and $|\varpi |=q^{-1}$, the difference is the compactly supported functionNormalize additive Haar measure by $\mathrm{vol}(\mathfrak{o})=1$. ThenThis difference survives in the Jacquet module. To prove that assertion, the unipotent action translates coordinates: $g_{\pi (n(y))f}(x)=g_f(x+y)$. Any such translation difference is compactly supported because the tail is $f(1)|x{|}^{-1}$ and $|x+y|=|x|$ when $|x|$ is sufficiently large. Its integral is zero: integrate over a sufficiently large additive ball, invariant under translation by $y$, and change variables. A finite sum of unipotent translation differences also has integral zero. Hence $h$, whose integral is nonzero, cannot be a coinvariant relation. In particular, on $J$ the operator $\pi (t)-{\chi }_0(t)I$ is nonzero. This proves the equal-character Jacquet self-extension for GL2 is nonsplit.

Any $T$-equivariant functional $J\to {\chi }_0$ must kill the nonzero image of that operator. Since $J$ has dimension two, its space of such functionals has dimension at most one; the quotient in the supplied exact sequence already provides one. Twisting back gives the same conclusion for the common linear character $\chi =\lambda$. For any different $\xi$, a surjection $J\to \xi$ is impossible because both composition factors of $J$ are $\chi$. Combining the two cases givesWhen the two listed linear characters coincide, the dimension remains one; they are not two independent contributions.