Here we suppose that $\Omega$ is a bounded subspace of $\mathbb{R}^n$ with $C^1$ boundary, denoted by $\partial \Omega$.
If $u\in H^1(\Omega)$ (with values in $\mathbb{R}^n$) then $(u,u)$ (scalar product in $\mathbb{R}^n$) belongs to $W^{1,1}(\Omega)$ (with values in $\mathbb{R}$).
It seems reasonable to have $|Tu|^2 = \tilde{T}(u,u)$ almost everywhere (on $\partial \Omega$) where $T : H^1(\Omega)\to L^2(\partial \Omega)$ and $\tilde{T} : W^{1,1}(\Omega)\to L^1(\partial \Omega)$ are the classical operators of traces and $|u|=\sqrt{(u,u)}$. Is it true ? If so, could you please give me the outline of the proof (or even a complete proof) or a reference.
First note that $$|Tu|^2=\tilde{T}(u,u)\tag{1},\ \forall\ u\in C^1(\overline{\Omega})$$
Now, take $u\in W^{1,2}(\Omega)$. Let $u_n\in C^1(\overline{\Omega})$ satisfies $\|u_n-u\|_{1,2}\to 0$. Note that $(u_n,u_n)\to (u,u)$ in $W^{1,1}(\Omega)$. By $(1)$ we must have $$\tag{2} |Tu_n|^2=\tilde{T}(u_n,u_n),\ \forall n$$
To proceed, first note that $Tu_n\to Tu$ in $L^2(\partial\Omega)$ and $\tilde{T}(u_n,u_n)\to \tilde{T}(u,u)$ in $L^1(\partial\Omega)$.
By taking subsequences (if necessary), we can assume further that $$Tu_n(x)\to Tu(x)\ \mbox{a.e. in}\ \partial\Omega$$
$$\tilde{T}u_n(x)\to \tilde{T}(u(x),u(x))\ \mbox{a.e. in}\ \partial\Omega$$
From the last convergences and $(2)$ we conclude that $$|Tu(x)|^2=\tilde{T}(u(x),u(x)),\ \mbox{a.e. in}\ \Omega$$
Remark: There are some gaps which need verification. I leave to you this task, however, if you need help, please leave a comment.