Here is a proposition in the book Neron Models giving various equivalent definitions of an unramified morphism of schemes.
In condition (e), why is finite put in parentheses? Normally when you state a theorem and write a property in parentheses, that is done to remind the reader that this property is always the case. But here, I see no reason why $\kappa(x)$ should be a finite extension of $\kappa(s)$.
In the affine case, if $\phi: A \rightarrow B$ is an algebra map of finite presentation (so $B$ is the quotient of a polynomial ring $A[t_1, … , t_n]$ by a finite generated ideal), $\mathfrak P$ is a prime of $B$, and $\mathfrak p = \phi^{-1}\mathfrak P$, there is no reason that the induced map
$$A_{\mathfrak p}/\mathfrak p A_{\mathfrak p} \rightarrow B_{\mathfrak P}/\mathfrak P B_{\mathfrak P}$$
be a finite extension of fields, or even a finitely generated extension of fields.
Best Answer
We'll show that any of the conditions that are not (e) imply that the given field extension is finite. We assume that the equivalence of (a)-(d) has already been proven.
Let $k=\kappa(s)$. By (d) and the fact that locally of finite presentation is stable under base change, we may assume $S=\operatorname{Spec} k$. By (c) and the fact that for a $k$-scheme $X$ the local dimension of $X\times_kX$ at $(x,x)$ is twice the local dimension of $X$ at $x$, we know that the local dimension of $X$ at $x$ must be zero, or in other words that $x$ is an isolated point of $X$. By $X\to S$ locally of finite presentation, we can pick an affine open neighborhood of $x$ of the form $\operatorname{Spec} k[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$, and since $x$ is an isolated point of $X$, we know that in fact $x$ is the spectrum of a field extension of $k$ of the form $k[y_1,\cdots,y_p]/(g_1,\cdots,g_q)$. In particular, this implies that $\kappa(x)$ is a finite extension of $k=\kappa(s)$.
On the other hand, one does need the finite extension property when proving the implication that (e) implies the other conditions - if we do not require it, the extension of fields $k\subset k(x)$ clearly has $\Omega_{k(x)/k}\neq 0$.