Let $G \hookrightarrow P \to M$ be a principal $G$ bundle, $P \times_\rho V$ be a vector bundle associated to representation $\rho$ of $G$ on $V$. If $\omega$ is a connection $1$-form on $P$ then we define for any $v \in C^\infty(M,P\times_\rho V)$ viewed as a function $v \in C^\infty(P,V)$ satisfying $v(p\cdot g) = \rho(g^{-1})v(p)$ the covariant derivative $$ \nabla^\omega v = dv + d\rho(\omega) \cdot v, $$ where $d\rho \colon \mathfrak g \to \mathrm{End}(V)$ and so $d\rho(\omega) \in \Omega^1(P,\mathrm{End}(V))$, and we extend this definition to the covariant exterior derivative $d^\omega$.
In one physical article I've faced with the following construction. If $g$ is a metric on $M$ we define the Hamiltonian by the formula $$ H_0 = - g \circ d^\omega \circ d^\omega, $$ where $d^\omega \circ d^\omega \colon C^\infty(M,P\times_\rho V) \to C^\infty(M,T^*M \otimes T^*M \otimes P \times_\rho V)$ and we use $g$ to contract "the $T^* M \otimes T^* M$ part". My question is the following. Actually $d^\omega \circ d^\omega v \in C^\infty(M,\Lambda^2 T^* M \otimes P \times_\rho V)$ for any $v \in C^\infty(M,P \times_\rho V)$. Doesn't this imply that $g \circ d^\omega \circ d^\omega$ must always give zero? This is because if $$ \theta = \sum_{i<j} b_{ij} dx^i \wedge dx^j = \sum_{i<j} \frac{b_{ij}}{2} dx^i \otimes dx^j - \sum_{i<j} \frac{b_{ij}}{2} dx^j \otimes dx^i, $$ then $$ g \circ \theta = \frac 1 2 \sum_{i<j} g^{ij} b_{ij} - \frac 1 2 \sum_{i<j} g^{ji} b_{ij} \overset{g^{ij}=g^{ji}}{=\!=\!=\!=} 0. $$