The problem is local, so we can assume $X$ affine. And since the equation for each added coordinate is independent of the rest, you get that your space is the fiber product over $X$ of the spaces defined by $O_X[T_i]/(T_i^n-x_i).$ (This is a very fundamental property of tensor products, which can be checked from the definition). If you believe what I said about ramification of fiber products, this reduces to checking what you said for a single $O_X[T_i]/(T^i-x_i). Now if a space is given by an equation like this over a base, the ramification divisor is the vanishing locus of the derivative.

Thanks. Could you say how Young symmetrizers give a basis? As far as I understand them, they project $V^{\otimes n}$ to the Schur functor, but they certainly don't project basis vectors to linearly independent elements. Is there some dual way of looking at them that fixes this problem?

Thanks Victor! These are indeed counterexamples to my (intended) question.

Thanks! The paper is exactly what I need. The trick seems to be to rotate a rectangle in the young tableau.