Natural map from spectrum of local ring (Lemma II 4.4 Hartshorne) Let $R$ be a valuation ring of a field $K$. Let $T=\operatorname{Spec} R$ and let $U=\operatorname{Spec}K$.In Harthorne Lemma II.4.4 he proves that given a scheme $X$, a morphism of $T$ to $X$ is equivalent to giving two points $x_0,x_1$ in $X$, with $x_0 \in \overline{\{x_1\}}$ and an inclusion $k(x_1)\subseteq K$, such that $R$ dominates the local ring $\mathcal{O}$ of $x_0$ on the subscheme $Z= \overline{\{x_1\}}$ of $X$ with its reduced induced structure.
Q1: The morphism $T \to X$ factors though $Z$: $T \to Z \to X$, and the first component of this induces two maps:
$$
O_{Z, x_0} \to O_{T, t_0} = R
$$
$$
O_{Z, x_1} \to O_{T, t_1} = K.
$$
From here it should follow that $O_{Z, x_0}$ dominates $R$:
that the first map is in fact an inclusion and that the 
$O_{Z, x_0} \cap m_{R} = m_{x_0}$, where $m_{x_0}$ is the unique maximal ideal of $O_{Z, x_0}$, and $m_R$ that of $R$.
Q2: In the converse direction, he states that the map $T \to X$ is obtained by  composing $\operatorname{Spec} \ R \to \operatorname{Spec} \ O_{Z, x_0} $, induced by $O_{Z, x_0} \to R$, 
with the natural map $\operatorname{Spec} \ O_{Z, x_0} \to X$. How is this natural map defined? Thank you.  

For the second question, $\mathcal{O}_{Z,x_0}$ can be considered as a local ring of some affine subscheme of $Z$, say $\operatorname{Spec}A$. Then its clear that $\mathcal{O}_{Z,x_0}=A_\mathfrak{p}$ where $\mathfrak{p}$ is the prime ideal corresponding to the point $x_0$. Since the open immersion $\operatorname{Spec}A\rightarrow Z$ has been defined, it suffices to give a homomorphism from $A$ to $A_\mathfrak{p}$, which is natural.
