About cotangent bundle $$\text{T}^*U\to\varphi(U)\times(\mathbb{R}^m)^*,\space(x,\lambda)\mapsto\left(\varphi(x),(D_{\varphi(x)}\varphi^{-1})^*(\lambda)\right)$$
What does $D_{\varphi(x)}\varphi^{-1}$ mean? Because I know "$D$" of sth, normally sth should be a function between two manifolds. But how can this homeomorphism work here?
 A: So this question is (I believe) about finding a functoriality for the cotangent bundle construction.
Because it is more familiar, let us first examine the functorial natural of the tangent bundle.  Let $\varphi:M\to N$ be a map of smooth manitolfds.  Then, given a point $x\in M$, we have a linear map $D\varphi_x: T_xM\to T_{\varphi(x)}N$, namely the derivative of $\varphi$ at $x$.  This can be computed by passing to a coordinate chart and then taking partial derivatives, if you so desire.  Just as we can glue the tangent spaces at each point into the tangent bundle, we can glue these linear maps together to define a map between tangent bundles:
$$D\varphi:TM\to TN, (x,v)\mapsto (\varphi(x),D\varphi_x(v))$$
Similarly, if we can use $\varphi$ to produce a linear map on the fibers of the cotangent bundle in a natural way, we can glue them together to form a map between the cotangent bundles.  Given a linear map $A:V\to W$, we have a dual map $A^*:W^*\to V^*$ by $A^*(f)(v)=f(A(v))$ for $f\in W^*$.  Unfortunately, because this reverses direction, we can't define our map as an ordered pair $(\varphi, (D\varphi)^*)$ 
However, if $\varphi$ is locally invertible near a point $x$ (equiv to $D\varphi_x$ being invertible), then we can correct for the fact that taking duals reverses direction by reversing a second time with inversion, namely we define a map 
$$(x,\lambda)\mapsto (\phi(x),((D\varphi_x)^*)^{-1}(\lambda)).$$
I assert that this is the same as the map you have.  To see this, we note that if $\varphi(x)=y$ and $\varphi$ is locally invertible at $x$, then $D(\varphi^{-1})_y=(D\varphi_x)^{-1}$.
So what is the term in the question?  It is the derivative of $\varphi^{-1}$ evaluated at $\varphi(x)$.  
Don't feel bad that you couldn't figure out what the notation it meant.  I had to come up with everything above before I could make sense of it. Parentheses around the $\varphi^{-1}$ would have made it significantly clearer to me.
