Let's first be clear about the map $\sigma$. By definition $\sigma_p:\mathfrak{g}\rightarrow \mathcal{V}_pP$, taking the lie algebra of $G$ into the vertical subspace of the tangent space of the total space $P$ at the point $p$. It's defined by $\sigma_p(X)=\displaystyle\frac{d}{dt}\bigg|_{t=0}p\cdot\exp{tX}$. An important fact about this map is that the connection form $\omega$, by definition, is a left inverse of $\sigma$: $\omega(\sigma(X))=X$
To use the formula
$d\omega(W_1,W_2)=W_1(\omega(W_2))-W_2(\omega(W_2))-\omega([W_1,W_2])$
it must be that $W_1, W_2$ are vector fields. So when you input the vertical vectors $(V_1)_p=\sigma_p(X), (V_2)_p=\sigma_p(Y)$ above, you first extend them to smooth vector fields whose value at $q\in P$ is $(V_1)_q=\displaystyle\frac{d}{dt}\bigg|_{t=0}q\cdot\exp{tX}=\sigma_q(X)$ and similarly for $V_2$. Thus, when you write $\sigma(X)$ this is a vector field, and the connection form $\omega$ maps this vector field to the constant vector $X\in \mathfrak{g}$.
It follows that
$V_1(\omega(V_2))=\sigma(X)(\omega(\sigma(Y)))=\sigma(X)(Y)$
i.e. the derivative of the constant vector $Y\in\mathfrak{g}$ in the direction $\sigma(X)$, which is $0$. That should address point (1) above.
With regard to point (2),
$[\sigma(X),\sigma(Y)]=\mathcal{L}_{\sigma(X)}\sigma(Y)$,
the derivative of $\sigma(Y)$ along the flow generated by the vector field $\sigma(X)$. But $\sigma(X)$ is defined as the derivative of the flow $(p,t)\mapsto p\cdot\exp(tX)=R_{\exp(tX)}(p)$. So
$(\mathcal{L}_{\sigma(X)}\sigma(Y))_p=\displaystyle\frac{d}{dt}\bigg|_{t=0}(R_{\exp(-tX)})_{*p\cdot\exp(tX)}(\sigma(Y)_{p\cdot\exp(tX)})=\displaystyle\frac{d}{dt}\bigg|_{t=0}(R_{\exp(-tX)})_{*p\cdot\exp(tX)}(\displaystyle\frac{d}{ds}\bigg|_{s=0}p\cdot\exp(tX)\exp(sY))=\displaystyle\frac{d}{dt}\bigg|_{t=0}\displaystyle\frac{d}{ds}\bigg|_{s=0}p\cdot\exp(tX)\exp(sY)\exp(-tX)=\displaystyle\frac{d}{dt}\bigg|_{t=0}\displaystyle\frac{d}{ds}\bigg|_{s=0}p\cdot\exp(sAd(\exp(tX))Y)=\displaystyle\frac{d}{dt}\bigg|_{t=0}\sigma(Ad(\exp(tX))Y)=\sigma(\displaystyle\frac{d}{dt}\bigg|_{t=0}(Ad(\exp(tX))Y))=\sigma([X,Y])$
