So I've spent quite a lot of time trying to work this proof out myself and I'm getting nowhere. The proof is Proposition 6.6 in Chapter II of Hartshorne. Let $X$ be a noetherian, integral, separated scheme which is regular in codimension $1$. Let $\mathbb{A}^{1}$ denote $\text{Spec } \mathbb{Z}[t]$. The proof is to show that $X \times \mathbb{A}^{1}$ (with the product taken over $\text{Spec }\mathbb{Z}$) is noetherian, integral, separated, and regular in codimension $1$. The noetherian, integral, and separated parts are easy if you know some basic change-of-base results. My question is about the regular in codimension $1$ part of the proof.
We can immediately reduce to the affine case. Let $X = \text{Spec }B$. Then the projection morphism takes the form $$ \pi: \text{Spec } B[t] \longrightarrow \text{Spec }B $$ corresponding to the inclusion of rings $$ B \hookrightarrow B[t] $$ If $p \in \text{Spec }B[t]$ is a point of codimension $1$, say it corresponds to a prime ideal $\mathfrak{p} \subset B[t]$ of height $1$. Hartshorne deduces that there are two cases:
1) $\pi(p)$ is a codimension $1$ point in $\text{Spec } B$;
2) $\pi(p)$ is the generic point of $\text{Spec } B$.
So far I understand. But this is as far as I understand. He claims that in the first case, $p$ is the generic point of $\pi^{-1}(p)$. But why is $\pi^{-1}(y)$ even a closed set, let alone irreducible? Moreover, why is it then immediate that $\mathcal{O}_{p} \simeq \mathcal{O}_{\pi(p)}[t]_{\mathfrak{m}_{\pi(p)}}$ as he claims?
He claims in the second case that the local ring $\mathcal{O}_{p}$ is a localization of $K[t]$ at some maximal ideal, where $K$ is the function field of $X$. This claim in itself was not clear to me, but I got an answer in another question here.
By exercise II.3.10 in Hartshorne, we have as topological spaces $$\pi|_{\operatorname{Spec}B[t]}^{-1}(y)\cong \operatorname{sp}\left(\operatorname{Spec}(B[t])_y\right)\cong \operatorname{sp}\left(\operatorname{Spec}(B[t]\otimes_B k(y))\right).$$ We notice that we have $B[t]\otimes_B k(y)\cong k(y)[t]$, and hence, $\pi|_{\operatorname{Spec}B[t]}^{-1}(y)$ is the spectrum of an integral domain. Thus it's an integral scheme and irreducible. In particular, it has a generic point $\zeta$. Notice that $\zeta$ is not the generic point of $\operatorname{Spec} B[t]$ as it is not sent to the generic point of $\operatorname{Spec} B.$ So, $\operatorname{codim}(\bar{\zeta},\operatorname{Spec} B[t])\ge 1.$ We also have that $\operatorname{codim}(\bar{p},\operatorname{Spec} B[t])= 1,$ and $\bar{p}\subset \bar{\zeta}$. But this implies that $\bar{p}=\bar{\zeta}.$
$\pi|_{\operatorname{Spec}B[t]}^{-1}(y)$ is endowed with the subspace topology; this implies that the closure of $p$ in $\pi|_{\operatorname{Spec}B[t]}^{-1}(y)$ is $\pi|_{\operatorname{Spec}B[t]}^{-1}(y)$. By the uniqueness of the generic point, we conclude that $p=\zeta$. This answers your first question.
For the second question, assume $y$ is represented by $\mathfrak{p}\le B.$ The extension of $\mathfrak{p}$ in $B[t ]$ is clearly given by $\mathfrak{p}[t]$. So, $p$ is an element of the closed set $V(\mathfrak{p}[t])$ by the properties of the extended ideal. We observe that $B[t]/\mathfrak{p}[t]\cong B/\mathfrak{p}[t]$ is a domain because $B/\mathfrak{p}$ is. Thus $\mathfrak{p}[t]$ is a prime ideal. It now follows that $\mathfrak{p}[t]$ is the generic point of $\pi|_{\operatorname{Spec}B[t]}^{-1}(y)$ and $p$ represents the prime ideal $\mathfrak{p}[t]$.
This shows that $\mathcal{O}_p=B[t]_{\mathfrak{p}[t]}$. To see the isomorphism, notice that we have a natural injective map $$B[t]\to B_{\mathfrak{p}}[t]\to \left(B_{\mathfrak{p}}[t]\right)_{\mathfrak{p}B_{\mathfrak{p}}[t]}.$$ This map induces the desired isomorphism $B[t]_{\mathfrak{p}[t]}\to \left(B_{\mathfrak{p}}[t]\right)_{\mathfrak{p}B_{\mathfrak{p}}[t]}\cong \mathcal{O}_{\pi(p)}[t]_{\mathfrak{m}_{\pi(p)}}.$