So Morse $f:M\to \mathbb{R}$ has nondegenerate critical point p iff $df|_{p}\pitchfork 0$-section.
Attempt
nondegenarate p iff Hessian has full rank at p iff
$Im(D|_{p}df)=T_{df(p)}^{*}M\Leftrightarrow Im(D|_{p}df)+T_{df(p)}$
$(0_{section})=T_{df(p)}^{*}M+T_{df(p)}M=T_{df(p)}(T^{*}M) \Leftrightarrow df \pitchfork 0_{section}$.
is this correct?
Is the tangent space at the 0-section zero dimensional since it is the constant map?
Things aren't quite right here, TKM. First, the Hessian $d^2f(p)$ is a symmetric bilinear map $T_pM\times T_pM\to\Bbb R$, so your image statement is not right. Second, you're talking about $df$ as a section of $T^*M$ and you want it to be transverse to the zero section of that bundle. Here's how you should approach it: Work in local coordinates, so assume $M=\Bbb R^n$ and $p=0$. Then the section corresponds to the function $\Bbb R^n\to \Bbb R^n\times\Bbb R^n$ given by $\Phi(x)=(x,df(x))$, i.e., $\Phi(x)=\big(x,(\partial f/\partial x_1,\dots,\partial f/\partial x_n)\big)$. To say $\Phi$ is transverse to $\Bbb R^n\times \{0\}$ at $0$ is to say $\text{im}(d\Phi_0) \oplus \Bbb R^n\times\{0\} = \Bbb R^n\times\Bbb R^n$, which is to say that the linear map associated to the bilinear form $d^2f(p)$ is surjective, which is precisely the definition of nondegeneracy.