Suppose I have a smooth manifold $M^n$ with tangent frame bundle $FM \xrightarrow{\pi} M$ with differential $\tau FM \xrightarrow {\pi_*} \tau M$. The kernel of $\pi_*$ is the vertical subbundle of the tangent bundle of $FM$, $V \subset \tau FM$, and $V$ is trivial, isomorphic over $FM$ to $FM \times \mathfrak{gl}_n$ via the map $FM \times \mathfrak{gl}_n \xrightarrow{\cong} V$ which gives the fundamental (vertical) vector field.
Suppose we are also given a connection form $\omega \in \Gamma(\operatorname{Hom}(\tau FM, FM \times \mathfrak{gl}_n))$.
On Wikipedia, it says that the torsion of the connection is a two-form with values in the tangent bundle $$T^\omega \in \Gamma(\operatorname{Hom}(\wedge^2 \tau M, \tau M))$$
which can be characterized either as the exterior derivative of the canonical one-form $$\text{Id} \in \Gamma(\operatorname{Hom}(\tau M, \tau M))$$ $$T^\omega = D^\omega(\text{Id}) \in \Gamma(\operatorname{Hom}(\wedge^2 \tau M, \tau M))$$ or as $$d\theta + \omega \wedge \theta$$
where $\theta$ is again the canonical one-form. (I deliberately didn't call it Id like I did the first time.)
Question: I'm not understanding this second characterization. How do I wedge $\theta$ with $\omega$?
Possible avenue:
Forms in $\Omega^k(M ; \tau M)$ are in bijection with tensorial forms in $\Omega^k(FM ; \pi^* \tau M)$, and $\pi^* \tau M $ is trivial, isomorphic over $FM$ to $FM \times \mathbb{R}^n$. So perhaps I have to make this translation for either $\omega$ or for $\theta$ to get the appropriate meaning. But $\omega$ has values that are vertical tangent vectors in $\tau FM$, not elements of $\pi^*\tau M$, so I'm not sure how to make this translation for $\omega$. If I make the translation for $\theta$, I get $\tilde{\theta} \in \Gamma(\operatorname{Hom}(\tau FM, \pi^*\tau M)) = \Gamma(\operatorname{Hom}(\tau FM, FM \times \mathbb{R}^n))$, which takes $v_U \in \tau_U FM$ to $(U, x)$, where $x \in \mathbb{R}^n$ is the expression of $\pi_*(v_U) \in \tau M$ in the basis given by $U^{-1}$. (The inverse may be wrong; it's just to make it a right action rather than a left action, or whatever.) But again, how do I wedge a $\mathfrak{gl}_n$-valued form with an $\mathbb{R}^n$-valued form to get a $\tau M$-valued form?
The notation $d\theta+\omega\wedge\theta$ is pretty condensed (and a bit problematic). The definition is that $(\omega\wedge\theta)(\xi,\eta)=\omega(\xi)(\theta(\eta))-\omega(\eta)(\theta(\xi))$, so you use that $\mathfrak{gl}(n,\mathbb R)=L(\mathbb R^n,\mathbb R^n)$. Using this definition, you can verify that $d\theta+\omega\wedge\theta$ is a horizontal, equivariant two-form with values in $\mathbb R^n$ and the space of such forms is isomorphic to the space of $\tau M$-valued two-forms on $M$.
A slightly less comprehensive common notation is that you write the components of $\theta$ as $\theta^i$ and the components of $\omega$ as $\omega^j_k$ and then define the torsion via $d\theta^i+\sum_j\omega^i_j\wedge\theta^j=\sum_{k,\ell}T^i_{k\ell}\theta^k\wedge\theta^\ell$. (To do this, you have to verify that the left hand side is horizontal.) Then you can compute that the $T^i_{k\ell}$ descend to a well defined tensor field on $M$.