Proof.
The problem is local on X and S. Hence we may assume that X = \mathop{\mathrm{Spec}}(B), S = \mathop{\mathrm{Spec}}(A) and f corresponds to a pseudo-coherent ring map A \to B.
If (1) holds, then B has finite tor dimension d as A-module. Then B_\mathfrak q has tor dimension d as an A_\mathfrak p-module for all primes \mathfrak q \subset B with \mathfrak p = A \cap \mathfrak q, see More on Algebra, Lemma 15.66.15. Then \mathcal{O}_ X has tor dimension d as a sheaf of f^{-1}\mathcal{O}_ S-modules by Cohomology, Lemma 20.48.5. Thus (1) implies (2).
By Cohomology, Lemma 20.48.5 (2) implies (3).
Assume (3). We cannot use More on Algebra, Lemma 15.66.15 to conclude as we are not given that the tor dimension of B_\mathfrak q over A_\mathfrak p is bounded independent of \mathfrak q. Choose a presentation A[x_1, \ldots , x_ n] \to B. Then B is pseudo-coherent as a A[x_1, \ldots , x_ n]-module. Let \mathfrak q \subset A[x_1, \ldots , x_ n] be a prime ideal lying over \mathfrak p \subset A. Then either B_\mathfrak q is zero or by assumption it has finite tor dimension as an A_\mathfrak p-module. Since the fibres of A \to A[x_1, \ldots , x_ n] have finite global dimension, we can apply More on Algebra, Lemma 15.77.5 to A_\mathfrak p \to A[x_1, \ldots , x_ n]_\mathfrak q to see that B_\mathfrak q is a perfect A[x_1, \ldots , x_ n]_\mathfrak q-module. Hence B is a perfect A[x_1, \ldots , x_ n]-module by More on Algebra, Lemma 15.77.3. Thus A \to B is a perfect ring map by definition.
\square
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