Let $f_0, f_1, g_0, g_1\sim\mathcal{N}(0, 1)$ i.i.d. Define
$$ \vec h = \begin{pmatrix} h_0\\ h_1 \end{pmatrix} = \begin{pmatrix} f_0g_0 + f_1g_1\\ f_0g_1 +f_1g_0 \end{pmatrix} $$
What is the characteristic function of $\vec h$?
Note that individually it is straightforward to verify that each $h_i$ is Laplace distributed, and these coordinates $\mathsf{Cov}(\vec h) = 2I_2$ are uncorrelated.
Laplace transform is, for $a,b$ small enough $$I=\mathbb{E}\left(e^{a(f_0g_0+f_1g_1)+b(f_0g_1+f_1g_0)}\right)=\mathbb{E}\left(e^{f_0(ag_0+bg_1)+f_1(ag_1+bg_0)}\right)= \mathbb{E}\left(e^{\frac{1}{2}[(a^2+b^2)(g_0^2+g_1^2)+4abg_0g_1]}\right)$$Let $$S=\left[\begin{array}{cc}a^2+b^2&2ab\\2ab&a^2+b^2\end{array}\right]=U^T\left[\begin{array}{cc}((a+b)^2)&0\\0&(a-b)^2\end{array}\right]U$$ where $U$ is an orthogonal matrix and denote $$\left(\begin{array}{c}g'_0\\g'_1\end{array}\right)=U\left(\begin{array}{c}g_0\\g_1\end{array}\right)$$ Observe that $g'_0$ and $g'_1$ are $N(0,1)$ and independent. As a consequence $$I=\mathbb{E}\left(e^{\frac{1}{2}[(a+b)^2(g'_0)^2+(a-b)^2(g'_1)^2]}\right)=\frac{1}{\sqrt{1-(a+b)^2}}\times \frac{1}{\sqrt{1-(a-b)^2}}$$
and the characteristic function is $$J=\mathbb{E}\left(e^{ia(f_0g_0+f_1g_1)+ib(f_0g_1+f_1g_0)}\right)=\frac{1}{\sqrt{1+(a+b)^2}}\times \frac{1}{\sqrt{1+(a-b)^2}}.$$ For instance if $b=0$ you get back the Laplace distribution that you mention.
Edit. More generally if $f$ and $g$ are independent and $N(0,I_d)$ and if $C_i$ are arbitary real square matrices of order $d$ then the characteristic function of $h_i=f^TC_ig$ for $i=1,\ldots,n$ is $$J=\frac{1}{\sqrt{\det(I_d+S(a))}}$$ where $$S(a)=(\sum_{i=1}^na_iC_i^T)(\sum_{i=1}^na_iC_i).$$