Prop 6.18,pg 64:The Thom class $\Phi$ on a rank $N$ oriented vector bundle $E$ can be uniquely characterized as the cohomology class in $H^n_{cv}(E)$ which restricts to the generator of $H^n_c(F)$ on each fiber $F$.
The proof wrote that
If $\Phi'$ is a coho. class in $H^n_{cv}(E)$ that restricts to a generator on each fiber then for $w \in H^n(M)$, $$ w \wedge \pi_* \Phi' = w. $$
I do not see how this holds from the definition of $\pi_*$
Definitions:
From Thom Isomorphism for an oriented bundle $E$ over $M$, we have isomorphism between de Rham coho. and compact vertical coho. $$ H^*(M) \rightarrow H^{*-n}_{cv}(E)$$ We define $\Phi$ the Thom class to be the image of $1$ in $H^0(M)$ under the above Thom isomorphism.
A form $w \in \Omega^*_{cv}(E)$ is locally of type (I) or (II), the map \begin{align*} (I) \quad & \pi^* \phi f(x, t_1, \ldots, t_n) dt_{i_1} \ldots d_{t_r} \mapsto 0 , \quad r< n \\ (II) \quad & \pi^* \phi f(x,t_1, \ldots, t_n) dt_1 \ldots dt_n \mapsto \phi \int_{\Bbb R^n} f(x,t_1, \ldots, t_n) dt_1 \ldots dt_n \end{align*} is our $\pi_* :\Omega^{*}_{cv}(E) \rightarrow \Omega^{*-n}(M)$. $E$ a vector bundle over $M$.