$\newcommand{\R}{\mathbb R}\newcommand\sgn{\operatorname{sgn}}\newcommand{\vpi}{\varphi}$Obviously, for $a:=\sqrt{2\pi}\,\psi(0)$ we have 
\begin{equation*}
	\psi(0)=f(0), 
\end{equation*}
where 
\begin{equation*}
	f:=a\vpi
\end{equation*}
and $\vpi$ is the standard normal density. Letting now $g(x):=\frac{\psi(x)-f(x)}x$ for $x\ne0$, with $g(0):=\psi'(0)$, we see that $g$ is a smooth integrable function and 
\begin{equation*}
	\psi(x)=f(x)+xg(x) \tag{1}\label{1}
\end{equation*}
for all real $x$. 

So, 
\begin{equation*}
	I(t)=\frac{J_f(t)+J_h(t)}{\sqrt t}, \tag{2}\label{2}
\end{equation*}
where $h(x):=xg(x)$, 
\begin{equation*}
J_f(t):=	
\int_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)}
\int_{\R}dx\,f(x)e^{ix^2y^2}, 
\end{equation*}
\begin{equation*}
	z_1:=e^{i\theta_1},\quad z_2:=e^{i\theta_2},
\end{equation*}
with $J_h(t)$ defined similarly. Here and in what follows, $t$ is any real number $>0$, unless specified otherwise. 

Note that 
\begin{equation*}
	\Im z_1\ne0\quad\text{and}\quad\Im z_2\ne0. \tag{3}\label{3}
\end{equation*}
Note also that 
\begin{equation*}
	\int_{\R}dx\,f(x)e^{ix^2y^2}=\frac a{\sqrt{1-2 i y^2}}
\end{equation*}
and hence 
\begin{equation*}
\begin{aligned}
J_f(t)&=	
a\int_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}}.  
\end{aligned}
\tag{4}\label{4}
\end{equation*}
For any fixed real $A>0$, by dominated convergence (which holds in view of \eqref{3}), 
\begin{equation*}
	\int_0^A\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}}
	\to\frac1{z_1z_2}\int_0^A\frac{dy}{\sqrt{1-2 i y^2}}\ll1  
\end{equation*}(as $t\to\infty$). 
We write $E\ll F$ to mean $|E|=O(F)$. 

So, letting $A$ go to $\infty$ slowly enough, we will have $A=o(\sqrt t)$ and 
\begin{equation*}
	\int_0^A\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}}
	=o(\ln t). 
\end{equation*}
Also, we will have 
\begin{equation*}
	\int_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}} \\ 
	\sim\int_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)y\sqrt{-2 i}}
\sim\frac{1+i}{4z_1z_2}\,\ln t. \tag{5}\label{5}
\end{equation*}
The latter asymptotic expression in \eqref{5} can be obtained by taking the latter integral in \eqref{5} in closed form, by partial fraction decomposition. It can also be obtained by writing $\int_A^{\sqrt t}=\int_A^{t^b}+\int_{t^b}^{t^{1/2}}$ for $b\in(0,1/2)$ such that $b$ is close to $1/2$. 

So, assuming $\psi(0)\ne0$, by \eqref{4}, 
\begin{equation*}
J_f(t)\sim
a\frac{1+i}{4z_1z_2}\,\ln t
=\frac{1+i}{4z_1z_2}\,\sqrt{2\pi}\,\psi(0)\ln t.  
\tag{6}\label{6}
\end{equation*}

Next, 
\begin{equation*}
J_h(t)=	\int_{\R}dx\,\sgn(x)g(x)K(t,x), \tag{7}\label{7}
\end{equation*}
where 
\begin{equation*}
	K(t,x):=\int_0^{\sqrt t}\frac{dy\,e^{ix^2y^2}|x|}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)}
	=\frac12\int_0^{r^2}dv\, e^{iv}H_r(v), 
\end{equation*}
\begin{equation*}
	r:=|x|\sqrt t, 
\end{equation*}
\begin{equation*}
	H_r(v):=\frac1{\sqrt v\,(\frac{\sqrt v}r-z_1)(\frac{\sqrt v}r-z_2)}. 
\end{equation*}
Further, 
\begin{equation*}
	2K(t,x)=K_1+K_2, \tag{8}\label{8}
\end{equation*}
where 
\begin{equation*}
	K_1:=\int_0^{1\wedge r^2}dv\, e^{iv}H_r(v),\quad K_2:=\int_{1\wedge r^2}^{r^2}dv\, e^{iv}H_r(v),
\end{equation*}
$u\wedge w:=\min(u,w)$. 

For $v\in(0,1\wedge r^2)$, we have $H_r(v)\ll\frac1{\sqrt v}$ and hence 
\begin{equation*}
	K_1\ll1. \tag{9}\label{9}
\end{equation*}
Note that $K_2=0$ if $r\le1$. If now $r>1$, then 
\begin{equation*}
	H'_r(v)=-\frac{H_r(v)}{2v}\,\Big(1+\frac1{\sqrt v-rz_1}+\frac1{\sqrt v-rz_2}\Big)
	\ll \frac{|H_r(v)|}v\ll\frac1{v^{3/2}}	
\end{equation*}
for $v>0$; so, integrating by parts, we see that  
\begin{equation*}
	K_2\ll1. \tag{10}\label{10} 
\end{equation*}

Collecting \eqref{7}--\eqref{10} and recalling that $g$ is an integrable function, we see that 
\begin{equation*}
J_h(t)\ll1. \tag{11}\label{11}
\end{equation*}

Collecting \eqref{2}, \eqref{6}, and \eqref{10}, we conclude that 
\begin{equation*}
I(t)\sim
\frac{1+i}{4z_1z_2}\,\sqrt{2\pi}\,\psi(0)\frac{\ln t}{\sqrt t},  \tag{12}\label{12}
\end{equation*}
as $t\to\infty$. 

---

As seen from the above proof, for \eqref{12} to hold it is enough that the function $\R\setminus\{0\}\ni x\mapsto\frac{\psi(x)-\psi(0)}x$ be integrable, with $\psi(0)\ne0$. 

---

Concerning the case $\psi(0)=0$: Then the bounds on $H_r(v)$ and $H'_r(v)$ developed above provide for dominated convergence. So, taking into account that for each real $x\ne0$ 
\begin{equation}
	H_r(v)\to\frac1{z_1z_2\sqrt v}, 
\end{equation}
we have 
\begin{equation}
	K(t,x)\to\frac1{2z_1z_2}\int_0^\infty\frac{dv\,e^{iv}}{\sqrt v}
	=\frac{1+i}{2z_1z_2}\,\sqrt{\frac\pi2}
\end{equation}
and 
\begin{equation*}
J_h(t)\to C_\psi:= c_\psi \frac{1+i}{2z_1z_2}\,\sqrt{\frac\pi2}, 
\end{equation*}
where 
\begin{equation}
	c_\psi:=\int_{\R}dx\,\sgn(x)g(x)=\int_{\R}dx\,\sgn(x)\frac{\psi(x)}x. 
\end{equation}
Therefore and because here $f=0$ and hence $J_f(t)=0$, we conclude that 
\begin{equation*}
I(t)\sim \frac{C_\psi}{\sqrt t}, 
\end{equation*}
as $t\to\infty$, provided that $\psi(0)=0$, $c_\psi\ne0$, and the function $g$ given by $g(x):=\psi(x)/x$ for $x\ne0$ is integrable. 

If $\psi(0)=0$ and $c_\psi=0$, then one has to dig deeper yet, possibly ad infinitum.