Heuristically, this result follows because the semiclassical wavefront set measures oscillations at frequency $\frac{\xi}{h}$. To understand this, observe that since $u$ does not depend on $h$, if $u$ is smooth, it has small high frequency oscillations, and hence, for $\xi\neq 0$, and hence points where $u$ is smooth and $\xi\neq 0$ do not appear in $WF_h(u)$. However, where $u$ is not smooth, $u$ has high frequency oscillations and so these points do appear.

Below you will find a proof of the statement.

To see that $WF_h(u)\cap \mathbb{R}^d\times \{0\}=$ supp $(u)\times\{0\}$ observe that
$$\mathcal{F}_h(\chi u)(0)=\langle u, \chi\rangle \neq 0$$
for $\chi$ supported near $x_0\in $ supp $(u)$ and
$$\mathcal{F}_h(\chi u)(0)=0 $$
for $\chi$ supported near $x_0\notin $ supp $(u)$.

To see the second part of $WF_h(u)$, simply use the characterization of $WF(u)$ that for $\xi\neq 0$, $(x_0,\xi_0)\notin WF(u)$ if and only if for $\chi$ supported sufficiently close to $x_0$, all $N>0$ and $\xi$ in a conic neighborhood of $\xi_0$,
$$|\mathcal{F}(\chi u)(\xi)|\leq C_N\langle \xi \rangle ^{-N}.$$

Thus, $(x_0,\xi_0)\notin WF(u)$ if and only if for all $N>0$ and all $\xi$ in a neighborhood of $\xi_0$,
$$|\mathcal{F}_h(\chi u)(\xi)|\leq C_N\langle \xi/h\rangle^{-N}.$$
But, since $\xi_0\neq 0$, this gives $(x_0,\xi_0)\notin WF(u)$ if and only if for all $N>0$ and $\xi$ in a neighborhood of $\xi_0$,
$$|\mathcal{F}_h(\chi u)(\xi)|\leq C_Nh^N$$

and hence if and only if $(x_0,\xi_0)\notin WF_h(u)$.