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Credit: This answer came out of trying to understand why auniket's answer (a.k.a. counterexample) works.

1) auniket is correct that for dimension reasons $T$ cannot surject onto $S$, so in particular my comment about $X$ being normal possibly helping is irrelevant. So is $T$.

2) It seems to me that there is a much more general problem with your desired statement. Namely I believe the following is true:

Claim: Under the conditions of the question, if in addition $\dim Y=0$ and $Y$ is reduced, then the desired statement cannot be true.

Proof: We may assume that $Y$ is a single point. Since by assumption $X$ is singular at $Y$, the local ring of $X$ at $Y$ is not a regular local ring. Therefore the ideal of $Y$ cannot be generated by $\dim X$ number of elements. On the other hand, by assumption $X$ is birational to $S$, so $\dim X=\dim S=s$. Therefore $v_1,\dots,v_s$ cannot generate the ideal of $Y$. $\square$

Note: I think this actually covers both of auniket's examples and would definitely give an arbitrary number of normal examples.

3) It seems that this still leaves a sliver of hope for you as your $Y$ is a curve (and even in the zero-dimensional case if $Y$ is non-reduced, it could work out). However, if it is reduced then you are at the absolute minimal number of generators that the singularity condition allows.

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Credit: This answer came out of trying to understand why auniket's answer (a.k.a. counterexample) works.

1) auniket is correct that for dimension reasons $T$ cannot surject onto $S$, so in particular my comment about $X$ being normal possibly helping is irrelevant. So is $T$.

2) It seems to me that there is a much more general problem with your desired statement. Namely I believe the following is true:

Claim: Under the conditions of the question, if in addition $\dim Y=0$ and $Y$ is reduced, then the desired statement cannot be true.

Proof: We may assume that $Y$ is a single point. Since by assumption $X$ is singular at $Y$, the local ring of $X$ at $Y$ is not a regular local ring. Therefore the ideal of $Y$ cannot be generated by $\dim X$ number of elements. On the other hand, by assumption $X$ is birational to $S$, so $\dim X=\dim S=s$. Therefore $v_1,\dots,v_s$ cannot generate the ideal of $Y$. $\square$

Note: I think this actually covers both of auniket's examples and would definitely give an arbitrary number of normal examples.

3) It seems that this still leaves a sliver of hope for you as your $Y$ is a curve (and even in the zero-dimensional case if $Y$ is non-reduced, it could work out).