Question about local description of the branch locus Let $\pi:Y\to X$ be a dominant, finite morphism of nonsingular varieties over an algebraically closed field $\Bbbk$. Assume furthermore that for all $Q\in Y$, with $P=\pi(Q)$, we have 
$$\mathcal O_{Y,Q}=\mathcal O_{X,P}[T_1,\ldots,T_k]/(T_1^n-x_1,\ldots,T_k^n-x_k)$$
for certain $x_i\in\mathcal O_{X,P}$. In other words, we have adjoined certain $n$-th roots. 
Now, let $R\subseteq Y$ be the ramification divisor and $H=\pi(R)$ the branch locus (both now with the reduced subscheme structure). Let $\mathcal I(H)$ be the ideal sheaf of $H$ and $\mathcal I(H)_P$ the stalk at $P$. Can I conclude that this ideal $\mathcal I(H)_P\subseteq \mathcal O_{X,P}$ is generated by the product $x_1\cdots x_k$? Or, correspondingly, can I conclude that $\mathcal I(R)_Q$ is generated by the product $y_1\cdots y_k$?
Edit. As pointed out in the answer by Dmitry Vaintrob, I want to assume $n$ not divisible by $\mathrm{char}(\Bbbk)$. Furthermore, we assume that the $x_i$ are reduced and coprime. 
 A: 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 - The fact that your ring is a fiber 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. Then the tensor product can be written in generators and relations as follows.
$$U\otimes_S V=S[x_1,\dots, x_n, x_1',\dots, x_m']/(R_1,\dots, R_k, R_1',\dots, R_l')$$ 
You can prove this in one line by using universal properties in the category of commutative $S$-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.
