Presumably in (2) you meant to assume the sets $A_p$ belong to $\mathcal U$.
The nontrivial thing is to show that (1) implies (2). The main point is that the Ramsey property of $\mathcal U$ implies that $\mathbb P$ has a $\mathcal U$-large suborder isomorphic either to $\omega$, $\omega^*$, or an infinite discrete order. Since (2) only depends on the $\mathcal U$-almost everywhere structure of $\mathbb P$, one only needs to check that (2) holds for these three orders. The case $\mathbb P = \omega$ is standard and the other two cases are basically trivial. Here are the details.
Enumerate $\mathbb P$ as $(p_n)_{n < \omega}$ and for $n < m$, set $f(p_n,p_m) = 0$ if $p_n < p_m$, $f(p_n,p_m) = 1$ if $p_n > p_m$, and $f(p_n,p_m) = 2$ if $p_n$ and $p_m$ are incomparable.
If $f$ has a $\mathcal U$-large 0-homogeneous set $H$, then you are essentially in the standard situation since $H \cong \omega$, so you finish using the standard argument.
If $f$ has a $\mathcal U$-large 1-homogeneous set $H$, then letting $n = \min\{k : p_k\in H\}$, $p_n$ is the maximum element of $H$. So if $(A_p)_{p\in \mathbb P}$ is a sequence with $A_p\in \mathcal U$ and $p < q$ implies $A_p\supseteq A_q$, then for $p < q$ in $H\cap A_{p_n}$, $q\in A_p$ since $p \leq p_n$ and hence $A_{p_n}\subseteq A_p$.
If $f$ has a $\mathcal U$-large 2-homogenous set, then there is a $\mathcal U$-large set of incomparables, and so (2) holds vacuously.
If I understand your question correctly, you ask if Defintion 7.37. and Definition 3. are equivalent for a Ramsey ultrafilter $\mathcal{U}(=\mathcal{H})$. The short answer is no, but let me elaborate.
Since for a $\mathcal{U}$-tree $T$, the set $[T]$ is technically a subset of $\omega^\omega$, we will only consider $\mathcal{U}$-trees $T$ such that $\text{stem}(T)$ is increasing. For such a tree $T$ there exists a pure refinement $T' \subseteq T$ such that every $x \in [T']$ is increasing, hence $\text{ran}(x) \in [\omega]^\omega$ can uniquely be identified with $x \in T'$.
Fact: If $\mathcal{U}$ is a Ramsey ultrafilter, then $\mathbb{M}_\mathcal{U}$ (Mathias forcing with an ultrafilter $\mathcal{U}$) is dense in $\mathbb{L}_\mathcal{U}$ (Laver forcing with an ultrafilter $\mathcal{U}$ such that for every $T \in \mathbb{L}_\mathcal{U}$ we have that $\text{stem}(T)$ is increasing). In particular, for every $T \in \mathbb{L}_\mathcal{U}$ there exists $A \in \mathcal{U}$ such that $\{\text{ran}(\text{stem}(T)) \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq [T]$.
Now Defintion 7.37. obviously implies Definition 3. : Let $\mathcal{X} \subseteq [\omega]^\omega$ be arbitrary such that $\mathcal{X}$ is $\mathcal{U}$-Ramsey. Let $T$ be a pure refinement of $\omega^\omega$ such that either $[T] \subseteq \mathcal{X}$ or $[T] \cap \mathcal{X} = \emptyset$. By the above Fact we can find $A \in \mathcal{U}$ such that $\{\emptyset \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq [T]$. Hence $\mathcal{X}$ is also $\mathcal{H}$-Ramsey.
On the other hand Definition 3. does not imply Defintion 7.37. : By an induction of length $\mathfrak{c}$ we can construct a $\mathcal{X} \subseteq [\omega]^\omega$ which is not $\mathcal{H}$-Ramsey. W.l.o.g. we can assume $0 \notin x$ for every $x \in \mathcal{X}$. Now define $\mathcal{X}':=\{\{0\} \cup x \,\, \colon \,\, x \in \mathcal{X}\}$, which is obviously $\mathcal{H}$-Ramsey. Now if $T$ is a $\mathcal{U}$-tree such that $\text{stem}(T)=\langle 0 \rangle$, there cannot exist a pure refinement $T' \subseteq T$ such that $[T'] \subseteq \mathcal{X}'$ or $[T'] \cap \mathcal{X}' = \emptyset$, because there does not exist an $A \in \mathcal{U}$ such that $\{\{0\} \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq \mathcal{X}'$ or $\{\{0\} \cup x \,\, \colon \,\ x \in [A]^\omega \} \cap \mathcal{X}' = \emptyset$.
Best Answer
See Section 12 in S. Todorcevic, Topics in Topology, Lecture notes in Mathematics Vol. 1652, Springer-Verlag Berlin Heidelberg 1997