About the second fundamental form
Solution 1:
I will use the notation $df_p$ rather than $\vec{\nabla}f(p)$.
First, we differentiate the function $N : p \mapsto \frac{df_p}{\|df_p\|}$ and get
$$dN_p(v) = \frac{H(f)_pv}{\|df_p\|} - \frac{df_p \cdot H(f)_pv}{\|df_p\|^3} df_p .$$
Therefore
$$\langle dN_p(v),w \rangle = \frac{\left ((df_p \cdot df_p) w - (df_p\cdot w)df_p\right)H(f)_pv}{\|df_p\|^3} \\ = \frac{\left (df_p \wedge(w \wedge df_p)\right)H(f)_pv}{\|df_p\|^3} = \frac{w H(f)_p v}{\|df_p\|}.$$
For your second question, first of all it is not difficult to see that $\phi_p$ is a second-degree polynomial.
Notice that by properties of second fundamental form, $\|df_p\|k_1$ and $\|df_p\|k_2$ are eigenvalues of $-H(f)_p$ with eigenvectors "principal direction" corresponding to $k_1$ and $k_2$ respectively, denoted $v_1$ and $v_2$. Thus it is clear that the vectors $(v_i,0)$ lie in the kernel of
$$\pmatrix{-H(f)_{(p)}-\|df_p\|k_i I_{3\times 3} & \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)& 0}.$$
Hence $\phi_p(\|df_p\|k_i) = 0$ for $i= 1,2$.
EDIT (to clarify the two points mentioned in your comments):
- $\phi_p$ is at most a third-degree polynomial. If we develop the determinant, the summand of $z^3$ is necessary the product of diagonal coefficients, which is zero.
- Firstly, $v_i$ is in the kernel of $-H(f)_{p}-\|df_p\|k_i I_{3\times 3}$. Since $v_i$ lies in the tangent space and $df_p$ is a normal vector, one also has $df_p\cdot v_i = 0$. So the product of this $4\times 4$ matrix by $(v_i,0)$ is zero.