Proof.
The equivalence of (1) and (2) is obvious as $\mathop{\mathrm{Spec}}(k)$ is a singleton and hence any morphism from it is quasi-compact.
Suppose $f$ is locally of finite type. Choose any affine open $\mathop{\mathrm{Spec}}(R) = U \subset S$ such that the image of $f$ is contained in $U$, and the ring map $R \to k$ is of finite type. Let $\mathfrak p \subset R$ be the kernel. Then $R/\mathfrak p \subset k$ is of finite type. By Algebra, Lemma 10.34.2 there exist a $\overline{f} \in R/\mathfrak p$ such that $(R/\mathfrak p)_{\overline{f}}$ is a field and $(R/\mathfrak p)_{\overline{f}} \to k$ is a finite field extension. If $f \in R$ is a lift of $\overline{f}$, then we see that $k$ is a finite $R_ f$-module. Thus (2) $\Rightarrow $ (3).
Suppose that $\mathop{\mathrm{Spec}}(R) = U \subset S$ is an affine open such that $f$ corresponds to a finite ring map $R \to k$. Then $f$ is locally of finite type by Lemma 29.15.2. Thus (3) $\Rightarrow $ (2).
Suppose $R \to k$ is finite. The image of $R \to k$ is a field over which $k$ is finite by Algebra, Lemma 10.36.18. Hence the kernel of $R \to k$ is a maximal ideal. Thus (3) $\Rightarrow $ (4).
The implication (4) $\Rightarrow $ (3) is immediate.
$\square$
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