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For the sake of simplicity, let $S=\operatorname{Spec}(k)$. I also suppose that there is no wild ramification, for instance requiring that $\textrm{char}(k) > \deg(f)$.

Answer to Question 1. It depends on the local behaviour of the cover around $R_f$. For instance, if $X$ is also smooth and $f$ is a Galois cover with group $G$, the multiplicity of each component $R_i$ of $R_f$ is $|\textrm{Stab}(R_i)|-1$, where $\textrm{Stab}(R_i)$ is the stabilizer subgroup of $R_i$.

Therefore, if $f$ is a double cover each component of $R_f$ appears with multiplicity $1$, if $f$ is a cyclic triple cover each component appears with multiplicity $2$ and so on.

Answer to Question 2. Yes, essentially by definition of canonical divisor. In particular, when $X$ is also smooth we have the identity $$K_X = f^*K_Y +R_f,$$ which is known as Hurwitz formula.

Answer to Question 3. Yes, $R_f$ is independent of $s$. In fact, its support coincides with the locus of points $x \in X$ where the differential $$df_x \colon T_xX \longrightarrow T_{f(x)}Y$$ is not an isomorphism, and this clearly does not depend on $s$.

For semplicitythe sake of simplicity, let $S=\operatorname{Spec}(k)$.

Answer to Question 1. It depends on the local behaviour of the cover around $R_f$. For instance, if $X$ is also smooth and $f$ is a Galois cover with group $G$, the multiplicity of each component $R_i$ of $R_f$ is $|\textrm{Stab}(R_i)|-1$, where $\textrm{Stab}(R_i)$ is the stabilizer subgroup of $R_i$.

Therefore, if $f$ is a double cover each component of $R_f$ appears with multiplicity $1$, if $f$ is a cyclic triple cover each component appears with multiplicity $2$ and so on.

Answer to Question 2. Yes, essentially by definition of canonical divisor. This In particular, when $X$ is also smooth we have the identity $$K_X = f^*K_Y +R_f,$$ which is known as Hurwitz formula.

Answer to Question 3. Yes, $R_f$ is independent of $s$. In fact, its support coincides with the locus of points $x \in X$ where the differential $$df_x \colon T_xX \longrightarrow T_{f(x)}Y$$ is not an isomorphism, and this clearly does not depend on $s$.

1

For semplicity, let $S=\operatorname{Spec}(k)$.

Answer to Question 1. It depends on the local behaviour of the cover around $R_f$. For instance, if $X$ is also smooth and $f$ is a Galois cover with group $G$, the multiplicity of each component $R_i$ of $R_f$ is $|\textrm{Stab}(R_i)|-1$, where $\textrm{Stab}(R_i)$ is the stabilizer subgroup of $R_i$.

Therefore, if $f$ is a double cover each component of $R_f$ appears with multiplicity $1$, if $f$ is a cyclic triple cover each component appears with multiplicity $2$ and so on.

Answer to Question 2. Yes, essentially by definition of canonical divisor. This is known as Hurwitz formula.

Answer to Question 3. Yes, $R_f$ is independent of $s$. In fact, its support coincides with the locus of points $x \in X$ where the differential $$df_x \colon T_xX \longrightarrow T_{f(x)}Y$$ is not an isomorphism, and this clearly does not depend on $s$.