The morphism $\phi:D\to X$ is going to factor through $Y\to X$.
Indeed you can reduce to $X = \operatorname{Spec}(A)$. Your morphism corresponds to $\phi^\sharp : A \to k[t]/(t^2)$. It maps $J$ into $(t)$ so it maps $J^2$ to $0$ and factors through $A\to A/J^2$.