What does it mean to say a boundary is $C^k$? I need a explanation on what does it mean to say a boundary is $C^k$.  Can anyone help me please.
And also need some explanation on how to straighten boundary ? 
 A: I don't know how much following will help you. But have it:
Let $M$ be any topological space such that each point $p\in M$ has a neighborhood $U\subset M$, homeomorphic to some open set of $[0,\infty)\times \mathbb R^{n-1}$ or open set of $\mathbb R^n$ via homeomorphism $\phi$.  We say that $(U,\phi)$ is a chart around $p$.  If every point has a such neighborhood, we say $M$ as topological manifold with boundary.  Now as  in wikipedia,(for terminology click on the link)
A smooth(C^k) manifold with boundary is a topological manifold with boundary equipped with an equivalence class of atlases whose transition maps are all smooth ($C^k$ ).
Difference with smooth manifold(C^k) and smooth manifold(C^k) with boundary is that for defining smooth manifold with boundary, we are allowing chart from open set of $[0,\infty)\times \mathbb R^{n-1}$ as well as open set of $\mathbb R^n$.
Here main point is that: If $p\in M$ has a neighborhood $U$ which is diffeomorphic ($\phi$) to some open set of $[0,\infty)\times \mathbb R^{n-1}$ such at $\phi(p)=0$, then there doesn't exists any coordinate neighborhood of $p$ which can be diffeomorphic to open set contained in interior of $[0,\infty)\times \mathbb R^{n-1}$.   Hence following definition makes sense:
Now define the boundary of smooth manifold with boundary as:
$$\partial M:=\{p\in M: \text{ there is a coordinate chart }(U,\phi)  \text{such that } \phi(p)=0  \}$$
We can prove that $\partial M$ is topological manifold of dimensional $n-1$.  If this manifold has differential structure of $C^k$, we say that boundary is of $C^k$.
