In general, you can say very little about the fibers of a normalization. It is possible that the total space is very nice yet the fibers have horrible singularities. (OK, not extremely horrible, but definitely not normal is not only possible, but the expected behavior for a special fiber).

In your case the situation is a little nicer, but based on only this much information one cannot tell what's happening. If $Z|_{t=0}$ is mock cover of ${\rm Spec} \left( A[[t]]/(t)\right)\simeq {\rm Spec}\, A$, then its irreducible components are isomorphic to ${\rm Spec}\, A$, so then your question depends on how these irreducible components come together.

The easiest case is if $Z|_{t=0}$ is a disjoint union of its irreducible components, in which case it is regular itself and since $Z|_{t=0}$ is a Cartier divisor in $Z$, it follows that then $Z$ is regular along $Z|_{t=0}$ and hence $\widetilde Z|_{t=0}=Z|_{t=0}$. However, this is a very special case.

It is more likely that the irreducible components of $Z|_{t=0}$ intersect, so it is itself not normal. Of course, this in itself is not necessarily a deal breaker, it just means that it is nearly impossible to tell what is happening.

The problem is that the normalization might separate some (or all) of these components, but which ones get separated and which don't depends on how the nearby fibers behave. I don't think you can give a simple criterion on what to expect.

So, the short answer is that based on the information given, what you hope for may or may not happen.

Example

For simplicity, let us assume that $A=k[x]$ for some field, although I am sure that the same construction can be done essentially for any $A$. Also, let us work over $A[t]$. Whatever we have can be specialized over $A[[t]]$ afterwards.

So then $A[t]=k[x,t]$ and hence ${\rm Spec}\, A[t]\simeq \mathbb A^2_k$. Next, let $X$ be a non-singular surface with a flat morphism $f:X\to \mathbb A^1_k\simeq {\rm Spec}\, k[t]$. Assume that $X_\lambda=f^{-1}((t-\lambda))$ is regular for $\lambda\neq 0$ and is a union of finitely many copies of $\mathbb A^1_k$ for $\lambda=0$. Assume further that $X$ admits two sections with images $C_1,C_2\subseteq X$ and let $Z$ be the surface obtained from $X$ by gluing $C_1$ to $C_2$ according to the identification of $C_1$ with $C_2$ given by $f$. It follows that $f$ factors through $Z$ and we still have a flat morphism $g:Z\to \mathbb A^1_k$. Now by Noether normalization $Z$ admits a finite morphism to $\mathbb A^2_k$ (I think one should be able to do this, so it is also Galois, but I haven't given this much thought, perhaps you can work this out). One can definitely make this so that it respects the morphism $g$: First take a finite map of $Z$ to $A^2_k$, then project $A^2_k$ to $A^1_k$ in a direction that no irreducible component of any fiber of $g$ is mapped to a point. Call the composition $h:Z\to A^1_k$ and take $p=g\times h: Z\to A^2_k$. This $p$ should satisfy the original conditions. The normalization of $Z$ is obviously $X$ and hence the normalization separates some components of $Z|_{t=0}$ while leaving some together. Varying this construction you can get all kinds of examples.

In general, you can say very little about the fibers of a normalization. It is possible that the total space is very nice yet the fibers have horrible singularities. (OK, not extremely horrible, but definitely not normal is not only possible, but the expected behavior for a special fiber).

In your case the situation is a little nicer, but based on only this much information one cannot tell what's happening. If $Z|_{t=0}$ is mock cover of ${\rm Spec} \left( A[[t]]/(t)\right)\simeq {\rm Spec}\, A$, then its irreducible components are isomorphic to ${\rm Spec}\, A$, so then your question depends on how these irreducible components come together.

The easiest case is if $Z|_{t=0}$ is a disjoint union of its irreducible components, in which case it is regular itself and since $Z|_{t=0}$ is a Cartier divisor in $Z$, it follows that then $Z$ is regular along $Z|_{t=0}$ and hence $\widetilde Z|_{t=0}=Z|_{t=0}$. However, this is a very special case.

It is more likely that the irreducible components of $Z|_{t=0}$ intersect, so it is itself not normal. Of course, this in itself is not necessarily a deal breaker, it just means that it is nearly impossible to tell what is happening.

The problem is that the normalization might separate some (or all) of these components, but which ones get separated and which don't depends on how the nearby fibers behave. I don't think you can give a simple criterion on what to expect.

So, the short answer is that based on the information given, what you hope for may or may not happen.

Example

For simplicity, let us assume that $A=k[x]$ for some field, although I am sure that the same construction can be done essentially for any $A$. Also, let us work over $A[t]$. Whatever we have can be specialized over $A[[t]]$ afterwards.

So then $A[t]=k[x,t]$ and hence ${\rm Spec}\, A[t]\simeq \mathbb A^2_k$. Next, let $X$ be a non-singular surface with a flat morphism $f:X\to \mathbb A^1_k\simeq {\rm Spec}\, k[t]$. Assume that $X_\lambda=f^{-1}((t-\lambda))$ is regular for $\lambda\neq 0$ and is a union of finitely many copies of $\mathbb A^1_k$ for $\lambda=0$. Assume further that $X$ admits two sections with images $C_1,C_2\subseteq X$ and let $Z$ be the surface obtained from $X$ by gluing $C_1$ to $C_2$ according to the identification of $C_1$ with $C_2$ given by $f$. It follows that $f$ factors through $Z$ and we still have a flat morphism $g:Z\to \mathbb A^1_k$. Now by Noether normalization $Z$ admits a finite morphism to $\mathbb A^2_k$ (I think one should be able to do this, so it is also Galois, but I haven't given this much thought, perhaps you can work this out). One can definitely make this so that it respects the morphism $g$: First take a finite map of $Z$ to $\mathbb A^2_k$, then project $\mathbb A^2_k$ to $\mathbb A^1_k$ in a direction that no irreducible component of any fiber of $g$ is mapped to a point. Call the composition $h:Z\to \mathbb A^1_k$ and take $p=g\times h: Z\to \mathbb A^2_k$. This $p$ should satisfy the original conditions. The normalization of $Z$ is obviously $X$ and hence the normalization separates some components of $Z|_{t=0}$ while leaving some together. Varying this construction you can get all kinds of examples.

In general, you can say very little about the fibers of a normalization. It is possible that the total space is very nice yet the fibers have horrible singularities. (OK, not extremely horrible, but definitely not normal is not only possible, but the expected behavior for a special fiber).

In your case the situation is a little nicer, but based on only this much information one cannot tell what's happening. If $Z|_{t=0}$ is mock cover of ${\rm Spec} \left( A[[t]]/(t)\right)\simeq {\rm Spec}\, A$, then its irreducible components are isomorphic to ${\rm Spec}\, A$, so then your question depends on how these irreducible components come together.

The easiest case is if $Z|_{t=0}$ is a disjoint union of its irreducible components, in which case it is regular itself and since $Z|_{t=0}$ is a Cartier divisor in $Z$, it follows that then $Z$ is regular along $Z|_{t=0}$ and hence $\widetilde Z|_{t=0}=Z|_{t=0}$. However, this is a very special case.

It is more likely that the irreducible components of $Z|_{t=0}$ intersect, so it is itself not normal. Of course, this in itself is not necessarily a deal breaker, it just means that it is nearly impossible to tell what is happening.

The problem is that the normalization might separate some (or all) of these components, but which ones get separated and which don't depends on how the nearby fibers behave. I don't think you can give a simple criterion on what to expect.

So, the short answer is that based on the information given, what you hope for may or may not happen.

In general, you can say very little about the fibers of a normalization. It is possible that the total space is very nice yet the fibers have horrible singularities. (OK, not extremely horrible, but definitely not normal is not only possible, but the expected behavior for a special fiber).

In your case the situation is a little nicer, but based on only this much information one cannot tell what's happening. If $Z|_{t=0}$ is mock cover of ${\rm Spec} \left( A[[t]]/(t)\right)\simeq {\rm Spec}\, A$, then its irreducible components are isomorphic to ${\rm Spec}\, A$, so then your question depends on how these irreducible components come together.

The easiest case is if $Z|_{t=0}$ is a disjoint union of its irreducible components, in which case it is regular itself and since $Z|_{t=0}$ is a Cartier divisor in $Z$, it follows that then $Z$ is regular along $Z|_{t=0}$ and hence $\widetilde Z|_{t=0}=Z|_{t=0}$. However, this is a very special case.

It is more likely that the irreducible components of $Z|_{t=0}$ intersect, so it is itself not normal. Of course, this in itself is not necessarily a deal breaker, it just means that it is nearly impossible to tell what is happening.

The problem is that the normalization might separate some (or all) of these components, but which ones get separated and which don't depends on how the nearby fibers behave. I don't think you can give a simple criterion on what to expect.

So, the short answer is that based on the information given, what you hope for may or may not happen.

Example

For simplicity, let us assume that $A=k[x]$ for some field, although I am sure that the same construction can be done essentially for any $A$. Also, let us work over $A[t]$. Whatever we have can be specialized over $A[[t]]$ afterwards.

So then $A[t]=k[x,t]$ and hence ${\rm Spec}\, A[t]\simeq \mathbb A^2_k$. Next, let $X$ be a non-singular surface with a flat morphism $f:X\to \mathbb A^1_k\simeq {\rm Spec}\, k[t]$. Assume that $X_\lambda=f^{-1}((t-\lambda))$ is regular for $\lambda\neq 0$ and is a union of finitely many copies of $\mathbb A^1_k$ for $\lambda=0$. Assume further that $X$ admits two sections with images $C_1,C_2\subseteq X$ and let $Z$ be the surface obtained from $X$ by gluing $C_1$ to $C_2$ according to the identification of $C_1$ with $C_2$ given by $f$. It follows that $f$ factors through $Z$ and we still have a flat morphism $g:Z\to \mathbb A^1_k$. Now by Noether normalization $Z$ admits a finite morphism to $\mathbb A^2_k$ (I think one should be able to do this, so it is also Galois, but I haven't given this much thought, perhaps you can work this out). One can definitely make this so that it respects the morphism $g$: First take a finite map of $Z$ to $A^2_k$, then project $A^2_k$ to $A^1_k$ in a direction that no irreducible component of any fiber of $g$ is mapped to a point. Call the composition $h:Z\to A^1_k$ and take $p=g\times h: Z\to A^2_k$. This $p$ should satisfy the original conditions. The normalization of $Z$ is obviously $X$ and hence the normalization separates some components of $Z|_{t=0}$ while leaving some together. Varying this construction you can get all kinds of examples.