On derivation of curvature formulas I can't figure out the idea behind the following part of the derivation of curvature formulas;
We let $M$ be a surface in $\mathbb{R}^3$ and $N$ be it's Gauss map. Moreover we consider a point $p\in M$ and local parametrisation $X$ such that $X(0)=p$.
Given this, there is a matrix  $A$ of $-dN: T_{p}M \rightarrow T_{p}M$ with respect to the basis $X_{u},X_{v}$. 
So far I understand things, now however, two matrix valued maps $DX=[X_{u},X_{v}]$ and $DN=[N_{u},N_{v}]$ are introduced from $U$ which is the domain of the parametrisation which satisfy ,
$-DN=DX A$
I dont understand this last equality.
The complete outline can be found on pages 46-47 in,
http://www.matematik.lu.se/matematiklu/personal/sigma/Gauss.pdf 
 A: If we understand where everything is coming from, this makes a lot more sense. Thus, I will try to explain the intuition behind each of these.
$[DX]: U\to \mathbb{R}^{3\times 2}$ is a map which goes from the 2 dimensional domain $U$ to two different vectors in $\mathbb{R}^3$ which together form the tangent space at that point on the manifold. 
$[DN]: U\to \mathbb{R}^{3\times 2}$ is a map which goes from the 2 dimensional domain $U$ to two different vectors in $\mathbb{R}^3$ which describe the changes in the normal vector as we travel along our domain $U$ before pushing it into $\mathbb{R}^3$.
$A: U\to \mathbb{R}^2$: I believe this is the function you are misunderstanding. As described in the linked PDF, it does not go from $T_pM\to T_pM$. It is still the matrix which represents how the normal changes, but instead, it is in the original domain, and is pushed into $\mathbb{R}^3$ by the function $X$ so we have
$$Ax=y \implies -dN(X(x)) = X(y)$$
Now, it should be easy to see that $-[DN]=[DX]\circ A$, as they are both performing the same operation. The introduction of the matrix $A$ just allows us to see that we can view this mapping of the derivative of the normal as a linear transformation in $U$ before mapping it to $\mathbb{R}^3$ via $X$.
