Assume the first three obstruction classes of a rank 4 vector bundle vanish and look at the fourth obstruction class. This fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (since $\pi\_3(SO\_4) \simeq \mathbb{Z} \oplus \mathbb{Z}$). Is there a geometric description of a system of generators in $\pi\_3(SO\_4)$ which is associated to these classes?


Recall that $SO\_4$ is double covered by $SU\_2 \times SU\_2$ and since 
$SU\_2 \cong S\_3$, $π\_3(SO\_4)=\pi\_3(S\_3) \oplus \pi\_3(S\_3)= \mathbb{Z} \oplus \mathbb{Z}$. The question is: how do the Euler and Pontryagin classes relate to this double cover? In other words, what is the system of generator $\langle \alpha, \beta \rangle$ of $\mathbb{Z} \oplus \mathbb{Z}$ so that given an element, if one writes it down as $a\alpha+b\beta$ then $a$ would be associated to the Euler class and $b$ to the Pontryagin class



Fourth obstruction, Pontryagin and Euler class
Geometric generators for $\pi\_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces an element of $\pi\_3(SO(4))$ denoted by $\alpha\_3$ and left multiplication induces an element denoted by $\beta\_3$. These generate $\pi\_3(SO(4))\cong\mathbb{Z}\oplus\mathbb{Z}$. 


The relation between obstruction classes and characteristic classes is discussed in 


* A. Dold and H. Whitney. Classification of oriented sphere bundles over a 4-complex. Ann. Math. 69 (1959), 667--677.


I think their Theorem 2 states that the part of the obstruction class corresponding to the generator $\beta\_3$ is exactly to the Euler class. On the other hand, the Pontryagin class of the bundle is $-4d\_1-2d\_2$ where $d\_1$ is the part of the obstruction class corresponding to $\alpha\_3$ and $d\_2$ is the part of the obstruction class corresponding to $\beta\_3$. Of course, then one can identify an actual element of the homotopy group corresponding to the Pontryagin class, but this will not be a generator of $\pi\_3(SO(4))$. Note that, contrary to what is implicitly claimed in the question, the obstruction class doesn't actually decompose as sum of Euler class and Pontryagin class (but this is consistent with the index of the Hurewicz map being 2). 


