Let $I \subset \mathbb{R}$ and $c:I \to M$ smooth. We define
$$\mathfrak{X}(M)_c := \Big\{X \in C^{\infty}(I,TM): X(t) \in T_{c(t)}M, \quad \forall t \in I\Big\}.$$ Then $X \in \mathfrak{X}(M)_c$ is called a vector field along the curve $c$.
Now, we get a unique map $$\frac{\nabla}{dt}: \mathfrak{X}(M)_c \to \mathfrak{X}(M)_c$$ defined by $$X \mapsto \frac{\nabla}{dt}X.$$
Now, one of the rules for this map that we have is that for $Z \in \mathfrak{X}(M)$, we get $$\frac{\nabla}{dt}(Z \circ c) = \nabla_{c'}Z$$ where $\nabla$ is the levi-civita connection (metric and torsion free connection). Now, in my notes, it says that $\nabla_{c'}Z$ is $C^{\infty}(I)$-linear,and I presume they mean with respect to the first argument, but I am unsure exactly what they mean. Since the connection is with respect to an underlying smooth manifold $M$ with metric $g$, i.e. SR-manifold $(M,g)$, we usually think of a connection (regardless of whether metric/torsion-free) as being $C^{\infty}(M)$-linear in the first argument. Can someone explain how to view this as a $C^{\infty}(I)$-linear map as opposed to $C^{\infty}(M)$-linear map? I did not understand the explanation in the notes.
Clarification: $C^{\infty}(M) := C^{\infty}(M,\mathbb{R})$ and $C^{\infty}(I) := C^{\infty}(I,\mathbb{R}).$
Edit: Here is the explanation given (somewhat paraphrased): The expression $\nabla_{c'}Z$ can be interpreted as an element in $\mathfrak{X}(M)_c$ in the following way. The map $$\mathfrak{X}(M) \times \mathfrak{X}(M) \to \mathfrak{X}(M)$$ given by $$(W,Z) \mapsto \nabla_{W}Z$$
is $C^{\infty}(M)$-linear in the first slot. Thus for fixed $Z \in \mathfrak{X}(M)$ the vector $\nabla_{W}Z(p) \in T_pM$ depends only on $W(p)$. Therefore, it makes sense to define $\nabla_{\nu}Z := \nabla_{W}Z$ where $W$ is any vector-field such that $W(p) = \nu$. As a consequence we obtain a well-defined linear map $$T_pM \to T_pM$$ $$\nu \mapsto \nabla_{\nu}Z$$ We may extend this to a $C^{\infty}(I)$-linear map
$$\mathfrak{X}(M)_c \to \mathfrak{X}(M)_c$$ $$X \mapsto \nabla_{X}Z$$ by setting $\nabla_{X}Z(t) = \nabla_{X(t)}Z \in T_{c(t)}M$. Setting $X = c'$ in the above, we see that $\nabla_{c'}Z$ is a $C^{\infty}(I)$-linear map.