So my question is as follows:
How does one show that if $M\subseteq \mathbb{R}^2$ is a manifold with positive gaussian curvature (G.C) at every point and if a curve $\gamma\subseteq M$ is such that $\gamma'\neq0$ for every $t$ then we can say that $\gamma''\neq0$?
I assumed WLOG that the curve is in natural parametrization at first and tried calculating the G.C via the principal directions, and Frenett-Serret equations but I got stuck.
How do I show that the second derivative is non-vanishing?
From your question it is not clear to me how much differential geometry of surfaces in $\mathbb{R}^3$ you already know. The reasoning below makes some assuptions in this regard, but I would not know, anyway, how to get this result without making use of these facts.
I will also assume that you meant to write that $M\subset \mathbb{R}^3$. You can then make use of the fact that, first of all, the Gauss curvature $K$ of $M$ in a given point is the product of the principal curvatures $k_1, k_2$ of $M$ in that point, so if $K>0$ then both principal curvatures are $\neq 0$ and they share the sign (depending on your choice of the normal to the surface).
On the other hand there is a well know relation between the principal curvatures $k_1, k_2$ of $M$ and the normal part $\kappa_n$ of the curvature of a curve $\gamma: (-a,a) \rightarrow M$ for which $\gamma^\prime \neq 0$, , which I will now explain:
If you assume that $\gamma$ is parametrized by arclength, then the curvature of $\gamma$ is just $||\gamma^{\prime\prime}||$ and you can decompose this into a tangential (wrt $M$) component $\kappa_g$ and normal component $\kappa_\nu$.
The index $g$ is related to the name of that component, which is usually called the geodesic curvature of $\gamma$ in $M$. The normal part is just $$\kappa_\nu=\langle\nu(\gamma(t)), \gamma^{\prime\prime}\rangle$$ with $\nu$ denoting the (a choice of a) normal of $M$ and clearly $\kappa_\nu\neq 0 $ implies $\gamma^{\prime \prime} \neq 0$, which is what you want to show.
Now, since $\gamma$ has its image in $M$, $\gamma^\prime$ is a tangent to $M$, so you know that $\langle \nu\circ \gamma, \gamma^\prime\rangle =0$ which implies, by differentiation $$ -\langle d\nu|_{\gamma} \gamma^\prime, \gamma^\prime \rangle = \langle \nu\circ\gamma, \gamma^{\prime\prime} \rangle$$ The term on the left hand side is just $$II(\gamma^\prime, \gamma^\prime) $$ the second fundamental form of $M$ evaluated at $(\gamma^\prime, \gamma^\prime)$ which you can express with the help of the principal curvatures of $M$ as $$II(\gamma^\prime, \gamma^\prime) = k_1 \cos(\phi)+ k_2\sin(\phi)$$ where $\phi$ is the angle between $\gamma^\prime $ and one of the principal directions. Since we noticed at the beginning, that both principal curvatures are nonzero, you see that $\kappa_n \neq 0$, which we saw implies $\gamma^{\prime\prime} \neq 0$