Let $(E, \pi, M)$ be a smooth vector bundle on a manifold. A connection is a map $\nabla : \Gamma(E) \to \Gamma(E \otimes \Omega^1(M))$ that satisfies certain properties (linearity and Leibnitz rule). Using this, we can construct, for each vector field $X$ on $M$, a map $\nabla_X: \Gamma(E) \to \Gamma(E)$.
Now let $\gamma:(0,1) \to M$ be a smooth curve. Several sources define a section $\sigma$ to be parallel if $\nabla_{\gamma'} \sigma = 0 $. How does this notion make sense? $\gamma'$ is not a vector field on $M$. At best, if $\gamma$ is injective, $\gamma'$ is a vector field on the image of $\gamma$.
When I asked about this, someone pointed me to the notion of a pullback connection. They said one can pull the connection back to a connection $\gamma^*\nabla$ on $(0,1)$ and ask if $(\gamma^*\nabla)_{\mathbf 1} (\gamma^*\sigma) = 0$. This doesn't solve anything. According to Wikipedia's definition of a connection pullback, we have $(\gamma^*\nabla) _\mathbf 1 (\gamma^*\sigma) = \nabla_{\gamma'}\sigma $, which takes us back to the original question.
I read through DoCarmo's book on Riemannian geometry and he seems to ignore this issue as well. Nor have I seen this question on StackExchange or Overflow. What am I missing?