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2 added 753 characters in body

How about we try the simple case $R=S$. R=S$and the identity map. Let$\mathfrak q$be a singular prime. The fibre over$\mathfrak q$is$\kappa (\mathfrak q)$which is regular (it is a field) but$R_q$R_{\mathfrak q}$ is not regular.

Edit: let me add a few more details since my original terse answer hardly deserves the upvotes! Let $P$ be the set of primes in $\text{Spec}(R)$ which contracts to $\mathfrak q$ (the set-theoretic fibre). Your first condition says that:

(1) $R_{\mathfrak p}$ is regular for all $\mathfrak p \in P$.

while the second says:

(2) $R_{\mathfrak p}/\mathfrak qR_{\mathfrak p}$ is regular for all $\mathfrak p \in P$.

So you can see where the problems come from: in general, for a local ring $R$, regularity of $R$ has nothing to do with regularity of $R/I$ for some ideal $I$, unless if $I$ is very special. That is how I think about the counter-examples, hopefully it helps.

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How about we try the simple case $R=S$. Let $q$ be a singular prime. The fibre over $q$ is $\kappa (q)$ which is regular (it is a field) but $R_q$ is not regular.