Chern-Weil: why do we divide by $2\pi$?

The inclusion $\mathbb{Z}\hookrightarrow\mathbb{R}$ induces a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow \check{H}^2(M;\mathbb{R})$ and under the identification $\check{H}^2(M;\mathbb{R})\cong H_{dR}^2(M;\mathbb{R})$ one has a homomorphism $\check{H}^2(M;\mathbb{Z})\longrightarrow H_{dR}^2(M;\mathbb{R})$.

I think that the problem lies in the use of THE for $\mathbb{Z}\hookrightarrow\mathbb{R}$. This is related to user72694's answer about the unit circle. If one is willing to define $S^1$ by $\mathbb{R}/2\pi\mathbb{Z}$, one can also define the inclusion $\mathbb{Z}\hookrightarrow\mathbb{R}$ by $\mathbb{Z}\ni k\mapsto 2\pi k\in\mathbb{R}$. This choice can wash $2\pi$'s from some formulae, and makes them appear elsewhere.

The point is to think of dual integer lattices. This can be seen already in the case of the unit circle. The class in cohomology dual to the fundamental homology class will be $d\theta$ divided by $2\pi$ since the circle has length $2\pi$. On the unit 2-sphere the curvature is $1$ but the total area is $4\pi$, which is why the fundamental cohomology class has to be normalized in a similar way to the circle to get an integer class. Hope this helps.