First Pontrjagin class and generator of $\pi_3(\mathrm{SO}(d))$ It is well-known that $H^4(B\mathrm{SO}(d), \mathbb{Z}) \cong \mathbb{Z}$, with a canonical generator given by $p_1$, the first universal Pontrjagin class.
Let's assume $d\geq 5$ so that everything is stable.
We have the following diagram of group homomorphisms:
$$H^4(B\mathrm{SO}(d), \mathbb{Z}) \stackrel{\alpha}{\longrightarrow} \mathrm{Hom}(\pi_4(B\mathrm{SO}(d)), \mathbb{Z}) \stackrel{\beta}{\longleftarrow} \mathrm{Hom}(\pi_3(\mathrm{SO}(d)), \mathbb{Z})$$
The left map is the usual map coming from the universal coefficient theorem and the Hurewicz homomorphism $\pi_4(B\mathrm{SO}(d) \to H_4(B\mathrm{SO}(d))$.
The right map is induced by the connecting homomorphism for the long exact sequence of homotopy groups for the classifying space fibration $\mathrm{SO}(d) \to E\mathrm{SO}(d) \to B\mathrm{SO}(d)$.
Now, there exists a generator $x$ of $\pi_3(\mathrm{SO}(d))$ such that the homomorphism $\varphi_x$ such that $\varphi_x(x) = 1$ satisfies $\alpha(p_1) = \beta(\varphi_x)$.
Q: What is an explicit description of this generator?
In particular, how does $x$ compare to the canonical map $y: \mathrm{SU}(2) \to \mathrm{SO}(4) \to \mathrm{SO}(d)$ for $d \geq 5$ (which, I think, is twice a generator)? Is $y = 2x$ or $y=-2x$? (Edit: As pointed out by Achim Krause, one has to fix an isomorphism $S^3 \cong \mathrm{SU}(2)$ to make this really well-defined).
Pontrjagin classes are commonly defined through Chern classes, which in turn are determined by their values on complex projective spaces.
My problem is that it seems quite involved to figure out what this really means in this concrete situation.
So what is an efficient way to determine the sign ambiguity?
 A: Let me try to summarize various observations from the comments and put it all together.
Let $d\ge 5$ and let $f: S^4\to BSO(d)$ be a generator of $\pi_4(BSO(d))\cong \mathbb Z$.
Let $\gamma^d$ be the universal orientable $\mathbb R^d$ vector bundle over $BSO(d)$.
Then we have the following  diagram extending the one by the OP. It is easily seen to commute by naturality.

Note that as indicated on the diagram  the bottom $\alpha$ and the right $f^*$ are isomormphisms.
Observe that $f^*(p_1(\gamma^d))=p_1(f^*(\gamma^d))$. It is well known that $p_1$ of any vector bundle over $S^4$ is divisible by 2 (if we identify $H^4(\mathbb S^4)$ with $\mathbb Z$)  and in fact every even number is realized. In particular  $p_1$ of the Hopf  $\mathbb R^4$ bundle is $\pm 2$. The above is done in detail for example in Milnor's classical paper on exotic $7$-spheres.
Since the two arrows ending in $Hom(\pi_4(\mathbb S^4),  \mathbb Z)$ are isomorphisms this implies that the image of the top row $\alpha$ is exactly $2\mathbb Z$ (we again identify $Hom(\pi_4(BSO(d)),\mathbb Z)$ with $\mathbb Z$).
Moreover this also shows that $\alpha(p_1)=\pm 2\beta(x)$ (recall that $x\in Hom(\pi_3(SO(d),\mathbb Z)$ is a generator).
Furthermore this shows that the clutching map of the stabilized Hopf bundle is exactly the generator of $\pi_3(SO(d))$ since if it wasn't then $p_1$ of the Hopf bundle would be a nontrivial multiple of $2$.
The clutching map of the Hopf bundle is just the identity map $\mathbb S^3=Sp(1)\to Sp(1)\cong SU(2)\subset SO(4)$ and therefore the generator of $\pi_3(SO(d))$ comes from the inclusion $SU(2)\to SO(4)\to SO(d)$.
This can also be seen in other ways without using Pontryagin classes. For example, recall the following well known picture of $SO(4)$.
Think of $\mathbb R^4$ as quaternions and $\mathbb S^3$ as $Sp(1)$ then consider the ineffective isometric linear  $\mathbb S^3\times \mathbb S^3$ action on $\mathbb R^4$ given by $(q_1,q_2)(v)=q_1vq_2^{-1}$. The kernel is $\mathbb Z_2=\{\pm (1,1)\}$ and we get $SO(4)=(\mathbb S^3\times \mathbb S^3)/\mathbb Z_2$. The first $\mathbb S^3$ commutes with multiplication by $i$ on the right and the second with multiplication by $i$ on the left. This identifies the two $Sp(1)$'s with two different embeddings of $SU(2)$ in $SO(4)$. The right $Sp(1)$ is easily seen to correspond to the standard embedding. The two $SU(2)$' give  generators of $\pi_3(SO(4))\cong\mathbb Z\oplus \mathbb Z$. Now look at the inclusion $SO(4)\to SO(5)$ and the induced map on $\pi_3$. From the homotopy sequence of the bundle $SO(4)\to SO(5)\to S^4$ we get that the map $\pi_3(SO(4))\to \pi_3(SO(5))\cong \mathbb Z$ is onto.
Since there is a unique irreducible real representation of $SU(2)$ of dimension 4 the two $SU(2)$'s are conjugate in $O(4)$ (but of course not in $SO(4)$). This conjugation can be made explicit by observing that $\overline{q(\bar v)}=v\bar q=vq^{-1}$. Hence quaternionic conjugation written as an element of $O(4)$ conjugates the two $SU(2)$'s.  Under the standard identification of $\mathbb R^4$ with $\mathbb H$ given by $(a,b,c,d)\mapsto a+bi+cj+dk$ the conjugating matrix is
diagonal $diag(1,-1,-1,-1)$.
The two $SU(2)$'s become conjugate in $SO(5)$ by $diag(1,-1,-1,-1, -1)$.
Hence the  $SU(2)$'s  have the same image in $\pi_3(SO(5))$ which must be the generator else $\pi_3(SO(4))\to \pi_3(SO(5))$ would not be onto.
Lastly, this could be seen using general theory by computing the Dynkin index of $SU(2)$ in $SO(5)$. Given a simple compact Lie group $G$ and its simple subgroup $H$ normalize their Killing forms $(\cdot, \cdot)_G$ and $(\cdot, \cdot)_H$ so that the longest roots have length $\sqrt 2$. Then the restriction of $(\cdot, \cdot)_G$ to the Lie algebra of $H$ is proportional to  $(\cdot, \cdot)_H$. The coefficient  is called the Dynkin index of $H$ in $G$.  It is always an integer and up to sign is equal to the coefficient in the induced map $\pi_3(H)\cong \mathbb Z\to \pi_3(G)\cong \mathbb Z$. For the proof see the book "Topology of transitive transformation groups" by Onishchik.
It is not hard to compute that the Dynkin index of $SU(2)$ in $SO(5)$ is 1 but won't do it as we have two other proofs already.
