Let $f: R\to S$ be a finite morphism of Commutative Noetherian rings, so $S$ is module finite over $R$ via $f$. Let $Q \in \text{Spec}(S)$, and set $P=f^{-1}(Q)$, so we have an induced map $R_P \to S_Q$. Set $\kappa(P):= \dfrac{ R_P}{PR_P}$. The degree of the induced finite morphism of schemes $f^*: \text{Spec}(S) \to \text{Spec}(R)$ of the point $P$ is given by $\dim_{\kappa(P)} \kappa(P) \otimes_{R_P} S_Q$. The scheme theoretic fiber of $P \in \text{Spec}(R)$ is given by $\kappa(P)\otimes_R S$. My question is:
Is it true that $\dim_{\kappa(P)} \kappa(P) \otimes_{R_P} S_Q = \dim_{\kappa(P)} \kappa(P)\otimes_R S$ ?
No, it is not true. Here is a simple explicit counterexample: let $k$ be a field, let $R=k$, let $S=k\times k$, and let $f:R\to S$ be given by $a\mapsto (a,a)$. Let $Q=k\times 0$, which makes $P=0$. Then $k(P)=k$, $R_P=R=k$, but $S_Q\cong k$ while $S=k^2$, so $\dim_{k(P)} k(P)\otimes_{R_P} S_Q = \dim_k k\otimes_k k = \dim_k k = 1$ while $\dim_{k(P)} k(P) \otimes_R S = \dim_k k\otimes_k k^2 = \dim_k k^2 = 2$.
For a conceptual explanation of what's going on here, the first dimension depends only on the point $Q$ while the second dimension depends on all the points in the fiber. So as soon as you have multiple points in your fiber, you should expect a difference.