I've read the proof of theorem 12.5, and in the last step, he said
... If the dimension $n$ is odd, then the Euler class itself has order $2$ by 9.4, so we have proved that $\mathfrak{o}_n(\xi) = e(\xi)$.
If the dimension $n$ is even, we must prove that $\lambda_n = +1$...
In here, $\xi$ is an arbitrary $n$-vector bundle over some manifold $B$ with fibre $F$.
But, we know that $\pi_{n-1}V_1(F)$ is same with infinite cyclic group if $n$ is odd and $\mathbb{Z}/2\mathbb{Z}$ if $n$ is even. (Note that, the top obstruction class $\mathfrak{o}_n(\xi)$ is in $H^n(B; \{ \pi_{n-1} V_1(F)\})$. It seems not to match what he want to prove and what we know.
What am I missing?
I'm not sure where you're getting your description of $\pi_{n-1}V_1(F)$: it is in fact always $\mathbb{Z}$. Indeed, $V_1(F)$ is nothing other than the unit sphere in $F$, which is $S^{n-1}$ since $F$ has dimension $n$. The chosen orientation on $\xi$ moreover gives an orientation on these unit spheres in all the fibers of $\xi$ at once, and so makes the local coefficients system $\{\pi_{n-1}V_1(F)\}$ just the trivial system of coefficients in $\mathbb{Z}$.
What is true is that if $n$ is odd, then the class $e(\xi)\in H^n(B;\mathbb{Z})$ satisfies $2e(\xi)=0$, and so $\lambda e(\xi)=e(\xi)$ for any odd integer $\lambda$. That's why the proof is complete in the case where $n$ is odd, having shown that $\lambda_n$ is odd.