Finite locally free equivalence quotients exist when orbits lie in affine opens

By [F1], cover $X$ by saturated affine opens $W_i$ and let $Y_i$ be their affine finite locally free quotients. For each intersection $W_i\cap W_j$, its image under $W_i\to Y_i$ is open, since finite locally free maps are open, and its inverse image is exactly the intersection by saturation. That open represents the quotient of the restricted relation, by [F2] and the local lifting/kernel-pair description. The corresponding open in $Y_j$ represents the same sheaf, so is uniquely isomorphic. These isomorphisms satisfy the cocycle identity by uniqueness; glue the $Y_i$ along them to a scheme $Y$.

The local quotient maps glue. Finite local freeness is local on the target, so $q:X\to Y$ is finite locally free and onto, and the local kernel-pair isomorphisms give $R=X{\times }_{Y}X$. Since every target point lifts after the covering $q$, and two lifts agree in the quotient precisely when related by that kernel pair, [F2] identifies $Y$ with the fppf quotient. A finite affine subcover of $X$ by the $W_i$ gives a finite affine cover of $Y$ by finite-type $k$-algebras, so $Y$ is finite type.

The image of the closed diagonal of $X$ under the finite closed map $q\times q$ is the diagonal of $Y$ as a subset, since $q$ is onto. Thus that diagonal has closed image. A finite-type $k$-scheme is locally separated: around every point of its diagonal, choose an affine open $T$ containing the point, and the restriction of the diagonal to $T\times T$ is closed. Hence its diagonal is an immersion; closed image makes this immersion closed, by checking the affine quotient ideals on those neighbourhoods and the open complement of the image. Therefore $Y$ is separated. AC is inherited from [F1]–[F2].