Why is Lemma 6.3 of Milnor's Lectures on the h-cobordism Theorem True?

I am reading Milnor's Lectures on h-cobordism Theorem right now, and let me explain here:

I prefer to write the diagram as

\begin{matrix} H_0(p)&\stackrel{j_*}{\longrightarrow}&H_0(N)\\ \downarrow{\psi}&&\downarrow{\psi}\\ H_r(\mathbb R^r,\mathbb R^r-\{0\})&\stackrel{i_*}{\rightarrow}&H_r(N\times\mathbb R^r,N\times(\mathbb R^r-\{0\}))\\ \downarrow{\cong}&&\downarrow{\cong}\\ H_r(U,U-p)&\stackrel{i_*}{\rightarrow}&H_r(V,V-N) \end{matrix}

where I assume $N$ has trivial normal bundle in $V$ and the indicated isomorphisms in the two columns are by excisions. We can certainly assume that the orientation on $\mathbb R^r$ is chosen so that the normal bundle $N\times \mathbb R^r$ has the same orientation as $V$, then the isomorphism $$H_r(N\times\mathbb R^r,N\times(\mathbb R^r-\{0\}))\xrightarrow{\cong}H_r(V,V-N)$$ send generator to generator. But now when you look at the left column isomorphism$$H_r(\mathbb R^r,\mathbb R^r-\{0\})\xrightarrow{\cong}H_r(U,U-p)$$ How does it send generator on $H_r(\mathbb R^r)$ to $H_r(U)$? It certainly depends on how we choose the orientation on $\mathbb R^r$. So by definition of intersection number, it is exactly sending generator on $H_r(\mathbb R^r)$ to $\epsilon\cdot$(generator of $H_r(U)$), and that's basically why the intersection number comes out here.