I'm studying a proof of the following:
Let $M$ be an oriented manifold of finite type without boundary. Let $R,S$ be two transversal embedded submanifolds without boundary; if $\eta_{R}, \eta_{S}, \eta_{R \cap S}$ denote respectively the compact Poincarè duals of $R,S, R \cap S$ in $M$, then:
$$ \eta_{R \cap S} = \eta_{R} \wedge \eta_S $$
By the tubolar neighborhood theorem, we can choose a tubolar neighborhood $T_R$ of $R$ such that the diffeomorphism between $T_R$ and $\mathcal{N}_{R|M}$ maps $T_R \cap S$ onto $(\mathcal{N}_{R|M})_{|R \cap S}$, the same holds for $S$. Set $T = T_R \cap T_S$. Let $\pi_R : T_R \mapsto R$ and $\pi_S : T_S \mapsto S$ be the bundle maps. By the transversality assumption, there is an isomorphism of vector bundles:
$$ \mathcal{N}_{R \cap S |M} \simeq (\mathcal{N}_{R|M})_{|R \cap S} \oplus (\mathcal{N}_{S|M})_{|R \cap S} \qquad (1) $$
Whence $T$ is a tubolar neighborhood of $R \cap S$ in $M$. We can choose representatives $\Phi_S, \Phi_R$ of the Thom classes of $\mathcal{N}_{S | M},\mathcal{N}_{R | M}$ such that: $$ (\Phi_R)_{|T}=\pi^{*}_{S}(\Phi_R)_{|S \cap T} \qquad (\Phi_S)_{|T}=\pi^{*}_{R}(\Phi_S)_{|R \cap T} \qquad (2)$$
A representative of the Thom calss of $\mathcal{N}_{R \cap S | M}$ is then:
$$\Phi(\mathcal{N}_{R \cap S |M})=\Phi((\mathcal{N}_{R|M})_{|R \cap S} \oplus (\mathcal{N}_{S|M})_{|R \cap S})=\pi^{*}_{S}(\Phi_R)_{|S \cap T} \wedge \pi^{*}_{R}(\Phi_S)_{|R \cap T}=\Phi_R \wedge \Phi_S $$
The conclusion now follows from the connection between the Poincarè dual of a submanifold and the Thom class of its normal bundle.
I basically have two problems: it is not clear, to me, why we are able to choose representatives of the Thom classes satisfying $(2)$; on the other hand, I cannot convince myself that such a choice is even necessary, can't we just consider isomorphism $(1)$ and "pass to Thom classes"? I'll be thankful for every suggestion or explanation.