Two finite disjoint petals can be made cross-incomparable

If $\mathrm{\varnothing }\in \mathcal{W}$, pair it with any other member; the conclusion is vacuous and there is another member because $\mathcal{W}$ is uncountable. Otherwise every member has positive size. Partition $\mathcal{W}$ by its finite sizes. F5 and A1 imply that some size $m>0$ occurs uncountably often; restrict to that subfamily, still denoted $\mathcal{W}$. Use AC to fix for each $S\in \mathcal{W}$ an enumeration $(x_i^S{)}_{i<m}$.

Fix an ultrafilter $\mathcal{U}$ on $\mathcal{W}$ as in F2. Suppose the conclusion fails. For $x\in T$ and $i<m$ let $Y(x,i)=\{R\in \mathcal{W}:x\text{ is comparable with }{x}_i^R\}$. For fixed $S\in \mathcal{W}$, failure supplies a comparable pair between $S$ and every $R\mathrm{\ne }S$, so the finite union of $Y(x,i)$ over $x\in S,i<m$ contains $\mathcal{W}\setminus \{S\}$. This cocountable set belongs to $\mathcal{U}$, and upward closure F6 puts the union in $\mathcal{U}$. F3 gives a pair $(x_S,i_S)$ with $x_S\in S$ and $Y(x_S,i_S)\in \mathcal{U}$. Use the already fixed finite enumeration and the usual order on $m\times m$ to take its first such pair.

Some $k<m$ occurs as $i_S$ on an uncountable subfamily $\mathcal{Z}$: otherwise its finitely many fibers would all be countable and F5 would make $\mathcal{W}$ countable. For distinct $S,R\in \mathcal{Z}$, F6 gives $V=Y(x_S,k)\cap Y(x_R,k)\in \mathcal{U}$, so F2 says $V$ is uncountable. Put $\gamma =\max \{\mathrm{ht}(x_S),\mathrm{ht}(x_R)\}+1<{\omega }_1$. The set $T_{<\gamma }$ is countable by F1 and F5, since $\gamma$ is countable. The map $H\mapsto x_k^H$ is injective on $\mathcal{W}$ because its members are disjoint. Consequently only countably many $H\in V$ can have $x_k^H\in T_{<\gamma }$. Choose an $H$ outside those exceptions. Its $k$th node is comparable with $x_S,x_R$ and strictly higher than both, so F4 forces it to extend both. Hence $x_S,x_R$ are comparable.

The set $C=\{x_S:S\in \mathcal{Z}\}$ is therefore an uncountable chain: injectivity follows from disjointness, and comparability from step 3.1. Its heights are unbounded in ${\omega }_1$, since any bounded collection of levels is countable by the same F1/F5 argument. Its downward closure $B=\{t:\mathrm{\exists }c\in C(t{\le }_{T}c)\}$ is a chain: compare two witnesses in $C$ and use F4 to compare both predecessors below the higher witness. It is cofinal. It is also maximal: if a node $u$ is comparable with every member of $B$, take $c\in C$ of height above $u$; F4 forces $u{<}_{T}c$, whence $u\in B$. Thus $B$ is a cofinal branch, contradicting F1. The failure assumed in step 2.1 is impossible, proving the assertion.