$\lambda\mapsto \sqrt{p}\lambda$, we obtain $$\begin{aligned}
&\frac{1}{ a }\sum_{\substack{m_1\,(a/p)\\m_2\,(p)}}
\sum_{\substack{\lambda\in {\mathcal{O}}_F/a\partial_F^{-1}\\ \lambda\lambda'\equiv m_1\,(a/p)}} \epsilon_p\left(\frac{-\lambda\lambda'+m_1}{a/p}+m_2\right)  e\left(\frac{{\operatorname{tr}}(\sqrt{p}\lambda\nu)-m_1 h}{ a}\right)e\left(-\frac{hm_2}{p}\right)\\
&=\frac{1}{ ap }\sum_{\substack{m_2\,(p)}}
\sum_{\substack{\lambda\in {\mathcal{O}}_F/a{\mathcal{O}}_F}}  e\left(\frac{{\operatorname{tr}}(\sqrt{p}\lambda\nu)-h\lambda\lambda'}{ a}\right)\epsilon_p\left(m_2\right) e\left(-\frac{hm_2}{p}\right)\\
&=\frac{1}{ a\sqrt{p} }\epsilon_p(h) \sum_{\substack{\lambda\in {\mathcal{O}}_F/a{\mathcal{O}}_F}}
e\left(\frac{{\operatorname{tr}}(\sqrt{p}\lambda\nu)-h\lambda\lambda'}{ a}\right) .\end{aligned}$$ Here, in the last line we have inserted the value of the Gauss sum. In particular we see that the latter quantity vanishes if $p\mid h$. If $p\nmid(h,a)$ we see by replacing $\lambda\mapsto \lambda+\frac{a}{r} \tau$ for $\tau\in {\mathcal{O}}_F$ that the latter quantity actually vanishes unless $(h,a)$ divides $\nu$. Therefore, as above, we put $r=(h,a)$ and write $h=h_1 r$, $a= a_1 r$, and $\nu=\nu_1 r$. Then the sum becomes $$\begin{aligned}
&\frac{r^2}{ a\sqrt{p} }\epsilon_p(h)
\sum_{\substack{\lambda\in {\mathcal{O}}_F/a_1{\mathcal{O}}_F}}  e\left(\frac{{\operatorname{tr}}(\sqrt{p}\lambda\nu_1)- h_1\lambda\lambda'}{ a_1}\right)\\
&=\frac{r}{ a_1\sqrt{p} }\epsilon_p(h)e\left(-\frac{\bar h_1 p \nu_1\nu_1'}{a_1}\right)
\sum_{\substack{\lambda\in {\mathcal{O}}_F/a_1{\mathcal{O}}_F}}  e\left(-\frac{h_1(\lambda-(\bar h_1\sqrt{p}\nu_1)')(\lambda'-(\bar h_1\sqrt{p}\nu_1)')'}{ a_1}\right)\\
&=\frac{r}{ a_1\sqrt{p} }\epsilon_p(h)e\left(-\frac{\bar h_1 p \nu\nu'/r^2}{a/r}\right) \sum_{\substack{\lambda\in
{\mathcal{O}}_F/a_1{\mathcal{O}}_F}}  e\left(-\frac{h_1\lambda\lambda'}{ a_1}\right).\end{aligned}$$ By [@Za] §4 Lemma 2 (noting that $p\nmid r$ implies $p\mid a_1$) we have $$\begin{aligned}
\frac{1}{ a_1 }\sum_{\substack{\lambda\in {\mathcal{O}}_F/a_1{\mathcal{O}}_F}} e\left(-\frac{h_1\lambda \lambda' }{a_1}\right) =
\epsilon_p( h_1)\sqrt{p}.\end{aligned}$$ We finally find that left hand side of is equal to $$\begin{aligned}
\epsilon_p(r) r \cdot e\left(-\frac{p \bar h_1 \nu\nu'/r^2}{a/r}\right).\end{aligned}$$ This concludes the proof of the Lemma.

We are now ready to compute the Fourier expansion of $\tilde\Phi_{mp^2}(z_1,z_2,s)$ in terms of the coefficients of the 