Consider $TM_p={(p,v;p\in M,v\in TM_p)}$ and let $M_0=\{(p,0;p\in M)\}$ be the zero section.
Let $X$ be a $C^r$ vector field($r\geqslant 1$) on a manifold $M$ and let $p_0\in M$ be a singularity of $X$.Then $p_o$ is simple singularity of $X$ if and only if the map $p\to(p,X(p))$ from $M$ to $TM$ is transversal(cross-sectional) to the zero section $M_0$ at $p_0$
Proof: Let $x:U\to\mathbb{R}$ be a local chart with $x(p_0)=0$. Let $TU=\{(p,v)\in TM;P\in U\}$. The map $Tx:TU\to\mathbb{R}^m\times\mathbb{R}^m$ defined by $Tx(p,v)=(x(p),Dx_p(v))$ is a local chart for $TM$.
Let $h=\pi_2 Tx X$, where $pi_2$ is the projection $\pi_2(x,y)=y$.Then $X$ is tranversal to $M_0$ at $p_0$ if and only if $p_0$ is a regular point of $h$ that is if $dh(p_0):TM_p\to\mathbb{R}^n$ is an isomorphism. On the other hand, $dh(p_0)=Dx(p_0)DX(p_0)$. Thus, $Dh(p_o)$ is an isomorphism if and only if $DX(p_0)$ is an isomorphism, which proves the proposition.
Question:
1)I am not understanding how the isomorphism relates to the regular point. If $DX(p_o)$ is an isomorphism implies $p_o$ is regular. Why? I thought that a regular point definition was: $p\in M$ then $X(p)\neq 0$. I do not see how this relates to the isomorphism $DX(p_0)$.
2) How is it possible to assure that $p\in M_0$, I mean a point in $M_0$ is singular? M_0 carries no information about $DX(p_0)$.
Thanks in advance!