Proof.
Openness of flatness shows (2) and (3) are equivalent, see Theorem 37.15.1.
Choose affine open U = \mathop{\mathrm{Spec}}(A) \subset X with x \in U and V = \mathop{\mathrm{Spec}}(R) \subset S with f(U) \subset V. Then R \to A is a flat ring map of finite presentation. Let \mathfrak p \subset A be the prime ideal corresponding to x. After replacing A by a principal localization we may assume there exists a quasi-finite map R[x_1, \ldots , x_ d] \to A, see Algebra, Lemma 10.125.2. Thus there exists at least one pair (U, g) consisting of an open neighbourhood U \subset X of x and a locally1 quasi-finite morphism g : U \to \mathbf{A}^ d_ S.
Claim: Given R \to A flat and of finite presentation, a prime \mathfrak p \subset A and \varphi : R[x_1, \ldots , x_ d] \to A quasi-finite at \mathfrak p we have: \mathop{\mathrm{Spec}}(\varphi ) is flat at \mathfrak p if and only if \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R) is Cohen-Macaulay at \mathfrak p. Namely, by Theorem 37.16.2 flatness may be checked on fibres. The same is true for being Cohen-Macaulay (as A is already assumed flat over R). Thus the claim follows from Algebra, Lemma 10.130.1.
The claim shows that (1) is equivalent to (4) and combined with the fact that we have constructed a suitable (U, g) in the second paragraph, the claim also shows that (1) is equivalent to (2).
\square
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