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show/hide this revision's text 3 some last-minute edits to part after "Edit".

There are two issues. You presumably mean that $x_i$ are themselves reduced and have no common divisors, or else the answer would be trivially no: just take $x_1=x_2$ and then the reduced branching ideal is generated by $x_1,$ not $x_1x_2.$ You also need to be careful what you mean in characteristic $p$, as then $T^p-x$ is everywhere ramified, no matter what $x$ is. Otherwise (in characteristic $0$ or relatively prime to $n$ and with $x_i$ reduced and having no common divisors) I don't see any reason why this would be false, just by writing out equations for the ramification and branching loci: ramification divisors under fiber products satisfy $R_{Y_1\times_X Y_2\to X}=R_{Y_1\to X}\times_X Y_2\cup Y_1\times R_{Y_2\to X}$, so it's enough to consider a single $T_1$ and you have $T_1^n-x_1=0, nT_1^{n-1}=0,$ which turns into $T_1=0, T_1^{n-1}=x_1$ (and the projection of this to $X$ is again $x_1=0$).

Is this what you were asking or am I missing something? If this is what you need, I don't deserve the bounty as I didn't say anything deep.

Edit: PS - To show what I said about The fact that your ring is a fiber products: suppose product follows from the following general statement. Suppose $U=S[x_1,\dots, x_n]/(R_1,\dots, R_k)$ and $V = S[x_1',\dots, x_m']/(R_1',\dots, R_l')$ are commutative algebras over $S$ given by generators and relations, where $R_i$ are relations on the $x$'s and $R_j'$ are relations on the $y$'s. A nice one-line proof of Then the fact that for a ring $S$, tensor product can be written in generators and relations as follows. $$U\otimes_S V=S[x_1,\dots, x_n, x_1',\dots, x_m']/(R_1,\dots, R_k, R_1',\dots, R_l')$$ is to use the You can prove this in one line by using universal properties in the category of commutative $S$-modules of a tensor product (a.k.a. coproduct in this category) and rings defined by generators & relations. S$-algebras. Namely, a map of $S$-algebras from the tensor product $U\otimes_S V$ to some $S$-algebra $W$ is equivalent to a pair of maps $f_1:U\to W, f_2:V\to W$ both over $S$, and a map from $S[x_1,\dots, x_n]/(R_1,\dots, R_k)$ to $W$ is equivalent to choosing a set of elements $w_1,\dots, w_n$ of $W$ satisfying relations $R_1,\dots, R_k$. Putting this together, $U\otimes_S V$ and the ring defined by the two sets of generators and relations as above satisfy the exact same universal property.
show/hide this revision's text 2 Added the explanation after "Edit", in response to request for clarification

There are two issues. You presumably mean that $x_i$ are themselves reduced and have no common divisors, or else the answer would be trivially no: just take $x_1=x_2$ and then the reduced branching ideal is generated by $x_1,$ not $x_1x_2.$ You also need to be careful what you mean in characteristic $p$, as then $T^p-x$ is everywhere ramified, no matter what $x$ is. Otherwise (in characteristic $0$ or relatively prime to $n$ and with $x_i$ reduced and having no common divisors) I don't see any reason why this would be false, just by writing out equations for the ramification and branching loci: ramification divisors under fiber products satisfy $R_{Y_1\times_X Y_2\to X}=R_{Y_1\to X}\times_X Y_2\cup Y_1\times R_{Y_2\to X}$, so it's enough to consider a single $T_1$ and you have $T_1^n-x_1=0, nT_1^{n-1}=0,$ which turns into $T_1=0, T_1^{n-1}=x_1$ (and the projection of this to $X$ is again $x_1=0$).

Is this what you were asking or am I missing something? If this is what you need, I don't deserve the bounty as I didn't say anything deep.

Edit: PS - To show what I said about fiber products: suppose $U=S[x_1,\dots, x_n]/(R_1,\dots, R_k)$ and $V = S[x_1',\dots, x_m']/(R_1',\dots, R_l')$, where $R_i$ are relations on the $x$'s and $R_j'$ are relations on the $y$'s. A nice one-line proof of the fact that for a ring $S$, $$U\otimes_S V=S[x_1,\dots, x_n, x_1',\dots, x_m']/(R_1,\dots, R_k, R_1',\dots, R_l')$$ is to use the universal properties in the category of $S$-modules of a tensor product (a.k.a. coproduct in this category) and rings defined by generators & relations. Namely, a map from the tensor product $U\otimes_S V$ to some $S$-algebra $W$ is equivalent to a pair of maps $f_1:U\to W, f_2:V\to W$ both over $S$, and a map from $S[x_1,\dots, x_n]/(R_1,\dots, R_k)$ to $W$ is equivalent to a set of elements $w_1,\dots, w_n$ of $W$ satisfying relations $R_1,\dots, R_k$. Putting this together, $U\otimes_S V$ and the ring defined by the two sets of generators and relations as above satisfy the exact same universal property.
show/hide this revision's text 1

There are two issues. You presumably mean that $x_i$ are themselves reduced and have no common divisors, or else the answer would be trivially no: just take $x_1=x_2$ and then the reduced branching ideal is generated by $x_1,$ not $x_1x_2.$ You also need to be careful what you mean in characteristic $p$, as then $T^p-x$ is everywhere ramified, no matter what $x$ is. Otherwise (in characteristic $0$ or relatively prime to $n$ and with $x_i$ reduced and having no common divisors) I don't see any reason why this would be false, just by writing out equations for the ramification and branching loci: ramification divisors under fiber products satisfy $R_{Y_1\times_X Y_2\to X}=R_{Y_1\to X}\times_X Y_2\cup Y_1\times R_{Y_2\to X}$, so it's enough to consider a single $T_1$ and you have $T_1^n-x_1=0, nT_1^{n-1}=0,$ which turns into $T_1=0, T_1^{n-1}=x_1$ (and the projection of this to $X$ is again $x_1=0$).

Is this what you were asking or am I missing something? If this is what you need, I don't deserve the bounty as I didn't say anything deep.