This is probably just a more streamlined version of the proof in the article you linked to, mentioning somewhat less differential forms. I don't know a more elementary proof, but hope this is of use anyway.

I'll write $D_t \subset \mathbb{R}^d$ for the domain of integration and I'll assume that the map $t\mapsto D_t$ is sufficiently regular in the sense that for sufficiently short times, the motion of the domain can be described by the flow of an ambient vector field $X$ on $\mathbb{R}^d$. I.e. for each $t\in \mathbb{R}$ there should exist a vector field $X$ on $\mathbb{R}^d$ with flow $\Phi_h:\mathbb{R}^d \to \mathbb{R}^d$ such that
$$
D_{t+h}=\Phi_{h}(D_{t}) \qquad \forall h \in (-\epsilon,\epsilon).
$$

(The field $X$ is not unique, one may add any vector field that's tangent to the boundary of $D_t$.)

With such a vector field $X$ given we compute
\begin{align}
\frac{d}{dt} \int_{D_t}F(\mathbf{x},t)d\mathbf{x}
&= \lim_{h\to 0}\frac{1}{h}\left( \int_{D_{t+h}}F(\mathbf{x},t+h)d\mathbf{x} - \int_{D_{t}}F(\mathbf{x},t)d\mathbf{x}\right)\\
&=\lim_{h\to 0}\frac{1}{h}\left( \int_{\Phi_h(D_{t})}F(\mathbf{x},t+h)d\mathbf{x} - \int_{D_{t}}F(\mathbf{x},t)d\mathbf{x}\right)\\
&=\lim_{h\to 0}\frac{1}{h}\left( \int_{D_{t}}\Phi^*_h(F(\mathbf{x},t+h)d\mathbf{x}) - \int_{D_{t}}F(\mathbf{x},t)d\mathbf{x}\right)\\
\end{align}

(The previous equality is just the substitution rule for integrals in several variables. So $\Phi^*_h\left(F(\mathbf{x},t+h)d\mathbf{x}\right)$ denotes the pullback of the differential form $F(\mathbf{x},t+h)d\mathbf{x}$ along $\Phi_h$. It's what you get by replacing every $\mathbf{x}$ with $\Phi_h(\mathbf{x})$ in the expression $F(\mathbf{x},t+h)d\mathbf{x}$. We continue by swapping limit and integral)

\begin{align}
&=\int_{D_{t}}\lim_{h\to 0} \frac{\Phi^*_h(F(\mathbf{x},t+h)d\mathbf{x}) - F(\mathbf{x},t)d\mathbf{x}}{h}
\end{align}

(To go on use Taylor $F(\mathbf{x},t+h)=F(\mathbf{x},t)+\frac{\partial F(\mathbf{x},t)}{\partial t}\cdot h+O(h^2)$ as well as linearity of $\Phi^*$ and $\Phi^*_0=\text{id}_{\mathbb{R}^n}$.)
$$
=\int_{D_{t}} \left(\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x}
+\lim_{h\to 0}\frac{\Phi^*_h(F(\mathbf{x},t)d\mathbf{x}) - F(\mathbf{x},t)d\mathbf{x}}{h}\right).
$$
Now that limit in the integral is per definition $L_X(F(\mathbf{x},t)d\mathbf{x})$, the Lie derivative of the form
$\omega=F(\mathbf{x},t)d\mathbf{x}$ with respect to $X$. Using Cartan's magic formula $L_x \omega = d(i_X \omega)+i_X(d\omega)$, which in our case simplifies to $L_x \omega = d(i_X \omega)$ since $F(\mathbf{x},t)d\mathbf{x}$ is of top degree, we get
$$
=\int_{D_{t}}\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x}
+\int_{D_{t}} d \left(i_X(F(\mathbf{x},t)d\mathbf{x})\right)
$$
Applying Stokes theorem to the second integral we finally arrive at
$$
\frac{d}{dt} \int_{D_t}F(\mathbf{x},t)d\mathbf{x}
=
\int_{D_{t}}\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x}
+\int_{\partial D_{t}} i_X(F(\mathbf{x},t)d\mathbf{x}).
$$
For those who want to get rid of that insertion in the second integral, a direct computation shows that $\int_\Gamma i_X( f(\mathbf{x})d\mathbf{x})=\int_\Gamma f(\mathbf{x}) X\cdot d\vec{A}$ where $d\vec{A}$ denotes the usual "surface element" with respect to the standard metric. So you can write the Leibniz integral rule as
$$
\frac{d}{dt} \int_{D_t}F(\mathbf{x},t)d\mathbf{x}
=
\int_{D_{t}}\frac{\partial F(\mathbf{x},t)}{\partial t}d\mathbf{x}
+\int_{\partial D_{t}} F(\mathbf{x},t)\, X\cdot d\vec{A}.
$$

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