Here's a toy model  that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$   then 

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc,  e^{-t\Lambda_n} x_n(0)\Big)$$

so 

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on  the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Alternatively we have $\newcommand{\bP}{\mathbb{P}}$
$$
\bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big]
$$
$$
=\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big]
$$