I am working on a proof of Cartan's formula $$ DA(u,v) = dA(u,v) + [A(u),A(v)], $$ where $A$ is a connection one form on a principal bundle and $u,v$ are two vector fields. $D$ denotes exterior covariant derivative One way of solving this should be using the identity $$ dA(u,v) = u(A(v)) - v(A(u)) - A([u,v]). $$ From the definition of the exterior covariant derivative, I am inclined to split my vector fields such that $u = H(u) + V(u)$, where H and V denote the horizontal and vertical components of the vector fields. In this way, I can use the fact that $A(H(u)) = 0$, but I am left with a lot of terms under the form such as $$v(A(V(u))).$$
My question is if there is any (defining) property of how a connection one form acts on a vertical vector field $V(u)$. My intuition/guess is that it should vanish or stay constant at least while going over a horizontal path, but I might be wrong.
Your formula $\require{enclose} \enclose{horizontalstrike}{is}$ [was] incorrect. Notice that if your [original] formula were true, we would have $DA=dA$ on coordinate vector fields, hence on all vector fields. The correct formula is $ DA(u,v) = dA(u,v) + [A(u),A(v)]. $
To prove the formula, we first check it when one of the vector fields is a fundamental vector field. Let $X \in \mathfrak g$ and let $X^\ast_p = \frac{\rm{d}}{\rm{d}t} p e^{tX} |_{t=0} $ be the corresponding fundamental vector field. We have $DA(X^\ast,v) = dA(X^\ast,v) + [A(X^\ast),A(v)] = - \mathrm{ad}_X A(v) + [X,A(v)] = 0$. Since an arbitrary vertical vector field can be expressed as a $C^\infty$ linear combination of fundamental vector fields, this shows that the formula holds when one of the vector fields is vertical.
Next we show that they have the same pullback under horizontal projection $H$: $ H^\ast DA = DA = H^\ast dA = H^\ast (dA+ \frac{1}{2}[A,A]) $ since $H^\ast [A,A] = [H^\ast A,H^\ast A]= 0$. (Perhaps confusingly, $[A,A](u,v) = [A(u),A(v)] - [A(v),A(u)] = 2 [A(u),A(v)]$.)
Now you can decompose $u$ and $v$ into their horizontal and vertical parts and see that the formula holds.