I must have mistunderstood something, this is giving me quite a headache. Please, do stop me once you notice an error in my thinking.
The Ehresmann Connection $v$ of some Bundle, $E\to M$, is the projection from $TE$ onto the Vertical Bundle. So $v \in \Omega^1(E,TE)$
One can define a Curvature form from this given by $\omega = dv + v\wedge v$, and have $\omega \in \Omega^2(E,TE)$.
Now, we can do this with the Tangent Bundle $TM\to M$ as well, and have for our Ehresmann connection $v\in \Omega^1(TM,TTM)$ and Curvature $\omega\in \Omega^2(TM,TTM)$.
According to http://en.wikipedia.org/wiki/Curvature_form , this curvature form $\omega$ should equal the Riemann Curvature Tensor $R$, $R(X,Y) = \omega(X,Y)$
But as I understand it, our Ehresmann Connection $v$ maps elements of $TTM$ back to $TTM$. And the Levi-Civita Connection is only interested in Elements of $TM$.
(1) how come the Curvature of the two agree? Aren't they something completely different?
Now I'm having the same problem with Chern classes. Chern Classes are given as Forms over $M$ ($\Omega(M,TM)$). But we can compute these Chern Classes via the Curvature form, which is a Form over $E$, $\Omega(E,TE)$ (or in our case $\Omega(TM,TTM)$, however, here I don't even understand how this can be in general).
(2) Again, aren't these two very different?