Firstly, a previous setting: For a differentiable map $f:\mathbb{R}^n \to \mathbb{R}^m$ its differential map on $p \in \mathbb{R}^n$ is $$d(f)_p: T_p(\mathbb{R}^n) \to T_{f(p)}(\mathbb{R}^m)$$ that goes through the respective tangent spaces. Since $T_q(\mathbb{R}^k)=\mathbb{R}^k$ we get $d(f)_p: \mathbb{R}^n \to \mathbb{R}^m$. Then the differential map takes vector $v \in \mathbb{R}^n$ and assign a vector $$d(f)_p(v) \in \mathbb{R}^n.$$
Now the question: Consider a curve $\alpha : (-\epsilon,\epsilon) \subset \mathbb{R} \to M$, where $M$ is a $m$-differential manifold with atlas $\{(U_i,\varphi_i)\}$.
Why the differential in $0 \in (-\epsilon,\epsilon)$ of the composition $\varphi \circ \alpha : (-\epsilon,\epsilon) \to \varphi(U) \subset \mathbb{R}^m$ is (directly) a vector? I mean, you have that $$d(\varphi \circ \alpha)_0 = (\varphi \circ \alpha)'(0)$$ is a (speed) vector in $\mathbb{R}^m$. Shouldn't be something as $$d(\varphi \circ \alpha)_0(v)$$ a vector? Is the reason that the curve only has $1$ parameter?
Thanks!
When talking about a function $\gamma:\mathbb{R}\to \mathbb{R}^n$, you have the identification $\gamma'(0)=d\gamma_0(1)$, where $1\in\mathbb{R}$ is considered as a tangent vector.
For example, for a function $f:\mathbb{R}\to\mathbb{R}$, you have $df_x:v\mapsto f'(x)\cdot v$, where you can read that $f'(x)=df_x(1)$.