$\newcommand{\R}{\mathbb R}\newcommand\sgn{\operatorname{sgn}}\newcommand{\vpi}{\varphi}$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 $\psi(0)\ne0$. (Of course, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
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\,\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 $g$ is integrable. (In the case when $\psi(0)=0$, we have $g(x)=\psi(x)/x$ for $x\ne0$. As was already noted, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
If $\psi(0)=0$ and $c_\psi=0$, then one has to dig deeper yet, possibly ad infinitum.
However, if $\psi$ is a "smooth bump", as you said in a comment, then apparently $\psi\ge0$ and hence $c_\psi>0$ (unless $\psi=0$ and hence $I(t)=0$).
Best Answer
First integrate over $\theta_1,\theta_2$. Use the delta function representation (for $k\in\mathbb{R}$) $$\int_{-\infty}^\infty e^{2\pi i k\theta}\,d\theta=\delta(k),$$ to evaluate $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{2\pi i (v_1\cdot x)\theta_1+2\pi i(v_2\cdot x)\theta_2}\,d\theta_1 d\theta_2=\delta(v_1\cdot x)\delta(v_2\cdot x).$$
Next for the integral over the vector $x\in\mathbb{R}^n$ use the coarea formula $$\int \delta\bigl(g(x)\bigr)f(x)\,d^nx=\int_{g(x)=0}\frac{f(x)}{|\nabla g(x)|}\,d^{n-1}\sigma(x),$$ once with $g(x)=v_1\cdot x$, $f(x)=\delta(v_2\cdot x)$, then once more with $g(x)=v_2\cdot x$, $f(x)=1$, to arrive at $$\int_{[-1,1]^n}\delta(v_1\cdot x)\delta(v_2\cdot x)\,d^nx=\frac{1}{|v_1||v_2|}\int_{[-1,1]^n\cap v_1\cdot x=0\cap v_2\cdot x=0}d^{n-2}\sigma(x).$$
So the integral in the OP equals the area of the $n-2$-dimensional region in $[-1,1]^n$ where $v_1\cdot x=0=v_2\cdot x$, divided by the product $|v_1||v_2|$ of the lengths of the two vectors $v_1,v_2$.
Dimensional check: if $x$ has dimension of length $L$ and $\theta_1,\theta_2$ are dimensionless, then the integral should have dimension $L^n$. The area of the $n-2$ dimensional region has dimension $L^{n-2}$, and since the product $|v_1||v_2|$ has dimension of $L^{-2}$ the dimensions match.