$M$ is is a Riemannian Manifold with an affine connection $\nabla$, $X$ is a vector field of which the restriction on the curve $\gamma$ is parallel. Fix some point $\gamma (t) $ on the curve. Is it true that for all $v\in T_{(\gamma(t))}M$, $\nabla_vX(\gamma (t))=0$?
By definition of parallel vector field along the curve, the covariant derivative of $X(\gamma(t))$ is 0. Would this imply the desire result?
Unless $X$ is very special, the answer will be no. To know the covariant derivative of $X$ with respect to your vector $v$, you don't just need to know the values of $X$ along any curve, but along a curve to which $v$ is tangent In particular, what you're asking will work if $v$ happens to be a multiple of $\gamma'(t)$. If this is not the case though, there may not be a curve $\alpha$ passing through $\gamma(t)$ to which $v$ is tangent so that $X$ is parallel along $\alpha$.
Drawing a picture in $\Bbb{R}^2$ might help see the idea. Start with a curve and a parallel vector field along that curve, and see if you can extend the vector field to something whose covariant derivative doesn't vanish anywhere but the curve.
