In the construction of exotic 7-spheres, Milnor used the fact the first Pontrjagin class $p_1(\xi_{ij})$ is linear in $i,j$. He claims that this is clear but I don't feel that. I tried to find some other notes about it. Then I found some more rigorous explanations, and they usually go as follows.
Recall that $\xi_{ij}$ is a rank 4 oriented vector bundle over $S^4$ obtained from the $[f_{ij}]\in \pi_3(SO(4)) \cong \mathbb Z \oplus \mathbb Z$, clutching function $f_{ij}(u)v= u^i \cdot v \cdot u^j$, where $\cdot$ stands for the quaternion multiplication.
The universal bundle $ESO(4) \rightarrow BSO(4)$ gives rise to an isomorphism $\pi_4(BSO(4)) \cong \pi_3(SO(4))$ (the connecting homomorphism $\delta$ of the homotopy l.e.s).
On the other hand, it can be verified that the map $\Theta: \pi_4(BSO(4)) \rightarrow H^4(S^4)$ defined by $[f] \mapsto p_1(f^* ESO(4))$ is actually a homomorphism. Therefore, by composing with the connecting map $\delta$, we obtain a homomorphism $$ \mathbb Z \oplus \mathbb Z \cong \pi_3(SO(4)) \rightarrow \pi_4(BSO(4)) \rightarrow H^4(S^4)$$
that sends $(i,j)$ to $p_1(\xi_{ij})$. It seems that we are done from here, but in this proof, we implicitly assumed two facts.
First, we used that the bundle obtained from the clutching function in $\pi_3(SO(4))$ and the bundle obtained from the pull-back of the universal bundle by the corresponding map of $\pi_4(BSO(4))$ (the preimage of the clutching function via the connecting homomorphism $\delta$) are the same, i.e. the following diagram
$$\require{AMScd} \begin{CD} \pi_4(BSO(4)) @>\delta>> \pi_3(SO(4)\\ @.{_{\rlap{f^*}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VVV\\ @. \mathrm{Vect}^{+}_{4}(S^4) \end{CD}$$
commutes.
Second, we also assumed that $[f_{i+i',j+j'}] = [f_{ij}] + [f_{i'j'}] \in \pi_3(SO(4))$.
Can someone tell me why these are true?