A geodesic on a Riemannian manifold $(M,g)$ with the Levi-Civita connection $\nabla$ is defined as a curve $\gamma(t)$ such that $ \nabla_{\gamma'(t)}\gamma'(t) \equiv0$.
Can we show that generally, the second fundamental form of a geodesic as a one-dimensional submanifold of $ (M,g) $ is $ 0 $?
I believe that the answer of Ernie060 is quite confusing as he is considering a curve in a surface in a 3-manifold. He is actually claiming that the geodesics of the surface are not geodesics of $\mathbb{R}^3$ which is clearly true for a generic surface. For those who are interested in minimal submanifolds, I would like to point out that these phenomena are very common. For instance, the Clifford torus is a minimal submanifold of the sphere but not of the euclidean space. (See DoCarmo Riemannian Geometry)
The answer to your question is actually yes.
Assume we have a general submanifold $\Sigma$ of $(M,g)$. The second fundamental form is defined as: $$ A(X,Y)=(\nabla_X Y)^N $$ with $X,Y \in T_p\Sigma$ and $\nabla$ Levi-Civita connection of M. If we consider $\Sigma$ one dimensional, (as $\gamma'(t)$ form a basis of $T_{\gamma (t)} \Sigma$ for all t) the second fundamental form is identically equal to zero by definition of geodesic.