Let $e_1,e_2$ be the principal directions, so that if $L$ denotes the Weingarten operator, we have $\left\langle Le_i,e_i\right\rangle=\kappa_i$, and write
$$e_{\theta}=\cos \theta e_1 + \sin \theta e_2 $$
Thus
$$k(\theta)= \left \langle Le_{\theta},e_{\theta}\right\rangle=\kappa_1\cos^2\theta +2\kappa_{12}\cos \theta \sin \theta + \kappa_2\sin^2\theta$$
Where $\kappa_{12}=\left\langle Le_1,e_2\right\rangle$. Integrating
$$\int_0^{2\pi}k(\theta)d\theta =\kappa_1\int_0^{2\pi}\cos^2\theta d\theta+2\kappa_{12}\int_0^{2\pi}\cos\theta\sin\theta d\theta+\kappa_2\int_0^{2\pi}\sin^2\theta d\theta =\pi(\kappa_1+\kappa_2) $$
(one can also prove that $\kappa_{12}=0$, but this is not needed here).
You cannot show that $S_p(Z)$ is zero, you have to show that $S_p(Z)$ is a multiple of $Z$, i.e. $S_p(Z) = \lambda(p,Z) Z$, where $\lambda$ is a function that depends on the direction $Z$ and the point $p$. Once you have this fact, you have to show that $\lambda$ does not depend on the direction $Z$ and that $\lambda$ is constant on the surface. Then you have proven that the surface is umbilical.
Hint: Consider a point $p$ on the surface. Show that, if the curvature of $\gamma$ is not $0$ at $p$, then the normal vector of $\gamma$ is equal to the surface normal.
This first hint should suffice as a starting point. If you need more input, leave a comment.
Addendum 1: further explanation. Your specific question (the case $\kappa(0)=0$) has already been asked before and a hint can be found in this question.
The hint actually shows that we can "work around" the directions in which the normal curvatures are zero.
A point is parabolic by definition if there is only one direction in which $\kappa_n$ is zero. A point is hyperbolic, by definition, if $K< 0$, or equivalently, if there are two directions in which the normal curvature is zero. So in case of a parabolic or hyperbolic point, we already know that in every direction, except in one or two directions, $S_p(Z) = \lambda(p,Z) Z$ holds.
Now take two linearly independent vectors $v$ and $w$, such that the normal sections in the directions $v$ and $w$ and $v+w$ are not zero. By the argument hereabove, we know that $S_p(v)=\lambda(p,v) v$, $S_p(w) = \lambda(p,w) w$ and $S_p(v+w) = \lambda(p,v+w) v+w$. Then by linearity
$$
\lambda(p,v+w)(v+w) = S_p(v+w) = S_p(v) + S_p(w)
= \lambda(p,v) v + \lambda(p,w) w.
$$
We may conclude that $\lambda(p,v)=\lambda(p)$ does not depend on the directions, except maybe in the one or two directions where the normal curvature is zero.
Consider now a direction $Z$ in which $\kappa_n = 0$. Write $Z=av+bw$ with $a,b\in \mathbb{R}$. Then by linearity
$$
S_p(Z) = a S_p(v) + b S_p(w) = a \lambda(p) v + b \lambda(p) w = \lambda(p) Z.
$$
So we conclude that the normal curvature section $\kappa_n$ must be equal in every direction, so the point is umbilical. Hence, in conclusion, we see that $p$ cannot be a parabolic or hyperbolic point.
Addendum 2: asymptotic directions. Consider a point at a surface. If $e_1$ and $e_2$ are the two principal directions with principal curvatures $\kappa_1$ and $\kappa_2$ respectively, then the normal curvature in the direction of $\cos \theta e_1 + \sin\theta e_2$ is
$$
\kappa_n = \kappa_1 \cos^2 \theta + \kappa_2 \sin^2 \theta.
$$
This is sometimes called Euler's formula. If $K = \kappa_1 \kappa_2$ is negative at the point, then $\kappa_1 > 0$ and $\kappa_2 < 0$. Then there are exactly two solutions $\theta$ such that $\kappa_n = 0$, which correspond to two linearly independent directions with zero normal curvature. These directions are called asymptotic directions. Also note that the point is planar (every normal curvature is zero) if and only if $\kappa_1=\kappa_2=0$.
Best Answer
You certainly do not want to conclude that $\kappa_1=\kappa_2=0$. No one said that the surface is planar. You have a typo in your formula for $\kappa_n = \langle \tilde N{}',\tilde T\rangle$.
This is the standard and important "trick" you should never forget: Since $\langle \tilde N,\tilde T \rangle = 0$ for all $s$, we get $$\langle \tilde N{}',\tilde T\rangle + \langle \tilde N,\tilde T{}' \rangle = 0,$$ so in this case — since $\gamma$ is a line — we have $$\kappa_n = \langle \tilde N{}',\tilde T\rangle = -\langle \tilde N,\tilde T{}' \rangle = -\langle \tilde N,0\rangle = 0.$$