# Covariant derivative of a vector field along a curve

In the following proposition of Do Carmo's Riemannian Geometry, I don't understand part (c) because $\frac{DV}{dt}$ is a vector field along the curve $c$, so it has domain $I$, while $\nabla_{\frac{dc}{dt}}Y$ is a vector field on $M$ with domain $M$, so how can they be equal? Also, the affine connection $\nabla$ is defined for two vector fields on $M$, $X$ and $Y$, and gives another vector field on $M$ $\nabla_X Y$. So how can we speak of $\nabla_{\frac{dc}{dt}}Y$ if $\frac{dc}{dt}$ is a vector field along $c$ and not a vector field on $M$? $$\frac{DV}{dt}(t_0) = \nabla_{\frac{dc}{dt}|_{t_0}}(Y) = \nabla_{\dot{c}(t_0)}(Y)$$
where both sides of the equality are elements of $T_{c(t_0)}M$.
Immediately after the statement of the theorem, Do Carmo proves that given an affine connection $\nabla$ on $M$, the value of $(\nabla_X Y)(p)$ depends only on $X(p)$ and the value of $Y$ along any curve $\alpha \colon I \rightarrow M$ satisfying $\alpha(0) = p$ and $\dot{\alpha}(0) = X(p)$. In particular, the expression $\nabla_{\frac{dc}{dt}|_{t_0}}(Y)$ makes sense (as we can interpret $\nabla_{\frac{dc}{dt}|_{t_0}}(Y)$ as $(\nabla_X Y)(c(t_0))$ where $X$ is any vector field on $M$ that satisfies $X(c(t_0)) = \frac{dc}{dt}|_{t_0}$ - such vector fields exist and $(\nabla_X Y)(c(t_0))$ does not depend on a specific choice of such an $X$).