I'm going to outline an argument, but it's sketchy at parts and I leave it to you to fill in the gaps. For convenience, I'm going to abbreviate $x^I := x^{i_1} \cdots x^{i_r}$ and $p_I = p_{i_1} \cdots p_{i_r}$.
Idea 1: If you squint at Theorems 12.4 and 12.9, you should believe that the class $x^I \in \pi_{4k} MSO$ is "dual" to the product of the Pontryagin classes $p_I$.
Idea 2: Recall the Pontryagin-Thom construction. In particular, recall how to go from an element $\pi_{4k} MSO$ to a (cobordism class of) a $4k$-manifold. Roughly an element $x \in \pi_{4k} MSO$ corresponds to a map $S^{4k+l} \to MSO(l)$ for $l \gg 0$. Since $S^{4k+l}$ is compact, this map factors through the Thom space of the tautological oriented $l$-plane bundle $\gamma_l^+$ over the oriented Grassmannian $\operatorname{Gr}^+(m,l)$ for $m \gg 0$. The preimage of the zero section $\operatorname{Gr}^+(m,l) \subset \operatorname{Th}(\gamma_l^+)$ in $S^{4k+l}$ is the wanted $4k$-manifold $M$. In other words, we have
\begin{array}{ccccc}
M & \rightarrow & S^{4k+l} \\
\downarrow & & \downarrow & \overset{x}{\searrow} \\
\operatorname{Gr}^+(m,l) & \xrightarrow[0]{} & \operatorname{Th}(\gamma_l^+) & \hookrightarrow & MSO(l)
\end{array}
where the left square is a pullback. By abuse of notation, I'm also going to call all the vertical maps $x$.
We want to compute the Pontryagin number of the stable normal bundle $\nu_M$. Since we've taken $l \gg 0$, I'm going to suppress mentioning "stable".
Idea 3: Identify this normal bundle using naturality. By construction, $\nu_M = (x^I)^* \nu_{\operatorname{Gr}}$ is the pullback of the normal bundle of the Grassmannian $\operatorname{Gr}^+(m,l)$ in $\operatorname{Th}(\gamma_l^+)$. This normal bundle is just $\gamma_l^+$.
The Pontryagin number we want to compute is $$\langle p_I(\nu_M), [M] \rangle = \langle (x^I)^* p_I(\gamma_l^+), [M] \rangle.$$ By definition of the Pontryagin classes using the universal bundle, $p_I(\gamma_l^+)$ is the cohomology class $p_I \in H^* \operatorname{Gr}^+(m,l)$.
But the conclusion of idea 1 is that $x^I$ is "dual" to $p_I$, which more precisely means in cohomology $(x^I)^* p_J$ is $[M]^\vee$ (dual to the fundamental class $[M] \in H_{4k} M$) if $I = J$, and $0$ otherwise.
So we have shown that $$\langle p_I(\nu_M), [M] \rangle = \langle \delta_{I = J} [M]^\vee, [M] \rangle = \delta_{I = J}.$$
The rest of the argument is straightforward.
