In the book Instantons and four-manifolds written by Uhlenbeck and Freedman, they say that:
We identify $\mathbb{R}^8$ as $\mathbb{H}^2$ with the standard inner product of $\mathbb{R}^8$, suppose $(q_1,q_2) \in S^7$,here $q_i$ lie in $\mathbb{H}$ satisfied $|q_1|^2+|q_2|^2=1$. Then $S^7 \to S^4:(q_1,q_2)\to [q1,q2]\in \mathbb{HP}\cong S^4$ defines a principal bundle with fiber $\mathrm{SU}(2)$. If we view $\mathfrak{su}(2)$ as $\mathrm{Im}\mathbb{H}$, then we can define connection form as $\omega:\mathrm{Im}(q_1d\bar{q}_1+q_2d\bar{q}_2)$ with vertical part $(gq_1,gq_2)$, here $g \in \mathrm{SU(2)}$.
I find that I can not check the property that $\mathrm{Im}(q_1d\bar{q}_1+q_2d\bar{q}_2)$ is identity on the vertical direction. I try to write vertical vector as $(\frac{d}{dt}g(t)q_1,\frac{d}{dt}g(t)q_2)$ but I have no idea to do more calculate, maybe I miss something fundamental.
For the second part of my question, consider the scaling map $\omega:\lambda:\mathbb{H} \to \mathbb{H}:x \to \lambda x$, here $\lambda$ a positive real number and a section $\mu$ on $\mathbb{H} \subseteq S^4 \cong \mathbb{HP}$, in coordinate is $[x,1]$.We pullback $\mathrm{Im}(q_1d\bar{q}_1+q_2d\bar{q}_2)$ by $\mu$ and obtain a connection $\mathrm{Im}(\frac{xd\bar{x}}{1+|x|^2})$ on $\mathbb{H}$. And the pullback connection $\mu^*\lambda^*\omega$ can be read as $\mathrm{Im}\frac{xd\bar{x}}{\lambda^2+|x|^2}$(Why?).
Since $\lambda$ not a map from $S^7$ to $S^7$, I think here they mean that pullback $\omega$ by the bundle map induced by $\lambda$. The map $\lambda:\mathbb{H}\to \mathbb{H}:x \to \lambda x$ and we can view it inside $\mathbb{HP}:[x,1] \to [\lambda x,1]$, this map induces the bundle map $S^7 \to S^7:(q_1,q_2)\to (\frac{\lambda q_1}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}},\frac{q_2}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}})$. Now for any tangent vector in $T_{(q_1,q_2)}P$, $\lambda^*\omega|_{(q_1,q_2)}(v)=\omega|_{(\frac{\lambda q_1}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}},\frac{q_2}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}})}(\lambda_* v)=\mathrm{Im}\frac{\lambda q_1}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}}d\bar{\frac{\lambda q_1}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}}}+\frac{q_2}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}} d\bar{\frac{q_2}{((\lambda q_1)^2+q_2^2)^{\frac{1}{2}}}}(\lambda_* v)$. But it's too complicated to compute, I think I must get in a wrong way.