Let $G = \operatorname{GL}_2(\mathbb Q_p)$, $K = \operatorname{GL}_2(\mathbb Z_p)$, and $B = TU$ the usual parabolic subgroup of $G$. For a pair of characters $\chi_1, \chi_2$ of $\mathbb Q_p^{\ast}$ and a complex number $s$, we may define the character $\chi_s$ of $B$ by $$\chi_s\begin{pmatrix} a & y \\ & b \end{pmatrix} = \chi_1(a)\chi_2(b)|\frac{a}{b}|^s.$$
We also have the modulus character $\delta_B \begin{pmatrix} a & y \\ & b \end{pmatrix} = |a/b|$.
We have the induced representation $V_s = \operatorname{Ind}_B^G \chi_s$, consisting of all locally constant functions $f: G \rightarrow \mathbb C$ satisfying $f(bg) = \chi_s(b)\delta_B(b)^{1/2}f(g)$ for all $b \in B, g \in G$.
By the Iwasawa decomposition $G = BK$, every $f \in V_s$ is completely determined by its restriction to $K$. It follows that if $\phi: K \rightarrow \mathbb C$ is locally constant, we can associate a flat section $f_{\phi}(g,s): G \times \mathbb C \rightarrow \mathbb C$ by
$$f_{\phi}(bk,s) = \chi_s(b) \phi(k).$$
A flat section is a special case of a holomorphic section, which is to say a function $f: G \times \mathbb C \rightarrow \mathbb C$ satisfying the following two properties:
For fixed $g \in G$, $s \mapsto f(g,s)$ is holomorphic.
For fixed $s \in \mathbb C$, $g \mapsto f(g,s)$ lies in $V_s$.
I am wondering in what sense, if any, are arbitrary holomorphic sections obtained from flat sections? If $f(g,s)$ is a holomorphic section, do there exist flat sections $f_1, ... , f_n$ and holomorphic functions $a_1, ... , a_n$ such that
$$f(g,s) = \sum\limits_{i=1}^n a_i(s)f_i(g,s)?$$
If we assume moreover that $f(g,s)$ is continuous on $G \times U$ then this is true. Consider the induced representation $\mathcal V = \operatorname{Ind}_{K \cap B}^K \chi_s$, which is independent of $s$. By the Iwasawa decomposition, the restriction map $f \mapsto f|_K$ is an isomorphism of vector spaces and of representations of $K$. The representation $\mathcal V$ of $K$ is preunitary, with pairing
$$\langle f_1, f_2 \rangle = \int\limits_{K \cap B \backslash K} f_1(k) \overline{f_2(k)}dk.$$
If we let $\phi_1, \phi_2, ...$ be an orthonormal basis of $\mathcal V$ with respect to this pairing, then for each fixed complex number $s_0$, and our holomorphic section $f(g,s)$, we have
$$f(k,s) = \sum\limits_{i=1}^{\infty} \Bigg( \int\limits_{K\cap B \backslash K} f(k',s)\overline{\phi_i(k')} dk'\Bigg) \phi_i(k) = \sum\limits_{i=1}^{\infty} a_i(s)\phi_i(k).$$ where each integral over $K \cap B \backslash K$ is holomorphic in $s$ (see note at the end), and for fixed $s$ the sum is finite (indeed, every element of $\mathcal V$ is a unique finite linear combination of the basis $\phi_i$). If we let $f_i(g,s)$ be the flat holomorphic section obtained from $\phi_i$, then we have
$$f(g,s) = \sum\limits_{i=1}^{\infty} a_i(s)f_i(g).$$
As we have mentioned, each $a_i(s)$ is holomorphic on $U$, and at each fixed $s \in U$, all but finitely many of the holomorphic functions $a_i$ take the value zero. This means that between the countably many holomorphic functions $a_1, a_2, a_3, ...$ there are uncountably many zeroes. The only way this can happen is there is an $N$ such that $a_n(s)$ is identically zero for all $n \geq N$.
Note: the holomorphicity of
$$a(s) = \int\limits_{K \cap B \backslash K} f(k,s)\overline{\phi_i(k)} dk$$
follows from the following result:
Suppose $(\Omega, dw)$ is a measure space, $U'$ is an open set in $\mathbb C$ and $h: \Omega \times U' \rightarrow \mathbb C$ is a measurable function such that $\int\limits_{\Omega} h(w,s)dw$ converges absolutely for each $s \in U'$, and such that $s \mapsto h(w,s)$ is holomorphic for each fixed $w \in \Omega$. If there exists an integrable function $H \geq 0$ on $\Omega$ such that $|h(w,s)| \leq H(w)$ for all $w, s$, then $s \mapsto \int\limits_{\Omega}h(w,s)dw$ is holomorphic on $U'$ with derivative $\int\limits_{\Omega} \frac{\partial h}{\partial s} (w,s)dw$.
If we assume that $f(g,s)$ is continuous, then on open precompact subsets $U'$ of $U$ the integrand $f(k,s)\overline{\phi_i(k)}$ is of course bounded by a constant. Our measure space $K \cap B \backslash K$ of course has finite volume, so we immediately get the holomorphicity of $a_i(s)$ by the mentioned result.
Of course, probably the full strength of that result is not needed, since our integrals are of course just finite sums.