Say $X=\mathbb A^n$ and $D_1,\dots, D_n$, the components of $D$, are the coordinate hyperplanes $x_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in etale topology.

$\pi_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of regular functions on $V$ is generated by the roots of monomials in $x$. You can write these as $x^m$ for some $m\in \mathbb Q^n$. The lattice $H$ generated by $m_i$ contains $\mathbb Z^n$, and the quotient $H/\mathbb Z^n$ is the dual abelian group $G^{\vee}$.

So what is $Y$ now? It is the normalization of the ring $k[x_1,\dots,x_n]$ in the bigger field $k(x^{m_i})$. So it is a toric problem now. The normalization is generated by monomials in the lattice $H$ which lie in the cone $(\mathbb R_{\ge0})^n$.

So $Y$ is toric, and every toric singularity is an abelian quotient singularity.

The condition $\pi_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this argument and the statement both fail in char $p$.

Say $X=\mathbb A^n$ and $D_1,\dots, D_n$, the components of $D$, are the coordinate hyperplanes $x_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in étale topology.

$\pi_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of regular functions on $V$ is generated by the roots of monomials in $x$. You can write these as $x^m$ for some $m\in \mathbb Q^n$. The lattice $H$ generated by $m_i$ contains $\mathbb Z^n$, and the quotient $H/\mathbb Z^n$ is the dual abelian group $G^{\vee}$.

So what is $Y$ now? It is the normalization of the ring $k[x_1,\dots,x_n]$ in the bigger field $k(x^{m_i})$. So it is a toric problem now. The normalization is generated by monomials in the lattice $H$ which lie in the cone $(\mathbb R_{\ge0})^n$.

So $Y$ is toric and simplicial, and every such singularity is an abelian quotient singularity.

The condition $\pi_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this argument and the statement both fail in char $p$.

Say $X=\mathbb A^n$ and $D_1,\dots, D_n$, the components of $D$, are the coordinate hyperplanes $x_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in etale topology.

$\pi_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of $Y$ is the normalization of $k[x_1,\dots,x_n]$ in the field $k(x^{m_1})$, where $m_i\in \mathbb Q^n$ generate the lattice $H$ with $\mathbb Z^n/H^*=G$ and $x^{m_i}$ are the corresponding Puisseaux (sp?) monomials.

So it is a toric problem now. The singularity is toric, hence abelian quotient.

The condition $\pi_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this does not work there.

Say $X=\mathbb A^n$ and $D_1,\dots, D_n$, the components of $D$, are the coordinate hyperplanes $x_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in etale topology.

$\pi_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of regular functions on $V$ is generated by the roots of monomials in $x$. You can write these as $x^m$ for some $m\in \mathbb Q^n$. The lattice $H$ generated by $m_i$ contains $\mathbb Z^n$, and the quotient $H/\mathbb Z^n$ is the dual abelian group $G^{\vee}$.

So what is $Y$ now? It is the normalization of the ring $k[x_1,\dots,x_n]$ in the bigger field $k(x^{m_i})$. So it is a toric problem now. The normalization is generated by monomials in the lattice $H$ which lie in the cone $(\mathbb R_{\ge0})^n$.

So $Y$ is toric, and every toric singularity is an abelian quotient singularity.

The condition $\pi_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this argument and the statement both fail in char $p$.