Let $X$ be a semi-normal projective variety and $p: \widetilde{X} \to X$ be the normalization. Suppose that $\widetilde{X}$ is smooth and there exists two smooth divisors $D_1, D_2 \subset \widetilde{X}$ such that $D_1 \cong D_2 \cong X_{\mathrm{sing}}$ and $p$ induces as isomorphism between $\widetilde{X} \backslash (D_1 \cup D_2)$ and $X\backslash X_{\mathrm{sing}}$. Note that, the non-singular divisors $D_1$ and $D_2$ map to $X_{\mathrm{sing}}$ under the morphism $p$. Can we conclude that $X$ has double point singularities along $X_{\mathrm{sing}}$?
Double Point Singularity in Semi-Normal Varieties – Algebraic Geometry
ag.algebraic-geometrynormalizationsingularity-theory
Related Solutions
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 $p$.
This works for any two closed sets. The relative cohomology of $X$ relative to a closed set $Z$ is the cohomology of $j_! \mathbb Q$ for $\mathbb Q$ the constant sheaf and $j$ the open immersion of $X\setminus Z$ into $X$. So if $Z_1$ and $Z_2$ are closed sets in $X$, $j_{12} \colon X \setminus(Z_1 \cup Z_2) \to X$ is the open immersion, $j_1 \colon X \setminus Z_1\to X$ is the open immersion, $j_1^2 \colon Z_2 \setminus (Z_1 \cap Z_2 ) \to Z_2$ is the open immersion, and $i_2 \colon Z_2 \to X$ is the closed immesion, then your claim follows from the short exact sequence of sheaves:
$$ 0 \to j_{12!} \mathbb Q \to j_{1!} \mathbb Q \to i_{2*} j_{1!}^2 \mathbb Q \to 0$$
which is possible to check: The first map is the adjunction map $j_{12!} j_{12}^* j_{1!} \mathbb Q\to j_{1!} \mathbb Q$, the second is the adjunction map $j_{1!} \mathbb Q\to i_{2*} i_2^* j_{1!} \mathbb Q$, and the exactness may be checked on stalks.
Best Answer
I think this is true if we assume that the isomorphisms $D_1 \cong X_{\mathrm{sing}} \cong D_2$ are also induced by $p$, at least in characteristic $0$ (though I think everything below is ok away from characteristic $2$). Edit: We also need to assume that the $D_i$ are disjoint in $\widetilde{X}$. If $D_1 \cup D_2$ is singular in $\widetilde{X}$ then we can also get worse singularities, see the edit below.
Let $\tau : D_1 \cong D_2$ denote the isomorphism induced by $p$ and let $\bar{X}$ be the quotient of $\widetilde{X}$ by the equivalence relation generated by $x \sim \tau(x)$. The quotient exists as a scheme in this case since $D_i \subset X$ are closed embeddings (this is an example of a pinching or Ferrand pushout, see e.g. this question and its answers). Then $\bar{X}$ is semi-normal with the required nodal singularities. Moreover, the map $p$ factors through a map $q : \bar{X} \to X$ by the universal property of quotients. By assumption $q$ is a bijection on points and isomorphism on residue fields so by semi-normality of $X$, $q$ is an isomorphism.
If we don't assume that the isomorphism $D_1 \cong X_{\mathrm{sing}} \cong D_2$ is induced by $p$, then this doesn't have to be true. For example, we can let $\widetilde{X}$ be two copies of $\mathbb{P}^2$ and $D_i$ be the conic $x^2 + y^2 + z^2$ the $i^{th}$ plane. Then we can consider the equivalence relation generated by identifying the two conics in the natural way as well as identifying $(x,y,z) \in D_i$ with $(y,x,z) \in D_i$. Then the quotient $X$ of $\widetilde{X}$ by this equivalence relation is semi-normal and $D_i \cong \mathbb{P}^1 \cong X_{\mathrm{sing}}$ but its singularities are not nodal. At a general point of $X_{\mathrm{sing}}$, the singularities will look like $\mathbb{A}^1$ times the union of the $4$ coordinate axes in $\mathbb{A}^4$ and $p|_{D_i}$ is $2$-to-$1$ onto $X_{\mathrm{sing}}$.
You might also want to take a look at Sections 5.1, 9.1 and 10.2 in Kollár's book Singularities of the minimal model program
Edit: Here is an example when $D_i$ are smooth but $D_1 \cup D_2$ is not. Let $D_i$ be the coordinate axes in $\mathbb{A}^2$ and consider the equivalence relation that identifies the two axes by swapping them. The resulting semi-normal surface $X$ is $\mathrm{Spec}$ of the the subring $$ \{f(x,y) \mid f(t,0) = f(0,t)\} \subset k[x,y]. $$ I think this is isomorphic to $u^3 - uvw + w^2 = 0$ in $\mathbb{A}^3$ which is not nodal. The singularities get worse if $D_1 \cup D_2$ has worse singularities, e.g. if $D_i$ meet in a tacnode.