Say $(\mathcal E, V)$ and $(\mathcal F, W)$ are Euclidean spaces. A map $F : \mathcal E \to \mathcal F$ is said to be differentiable at $p_0 \in \mathcal E$ if there exists a linear map $L : V \to W$ such that$$\lim_{p \to p_0} \frac{\lVert w - Lv \rVert}{\lVert v \rVert} = 0,$$where $v = p - p_0$ is the unique vector in $V$ satisfying $p = p_0+v$ and $w = f(p)-f(p_0)$ is the unique vector in $W$ satisfying $f(p) = f(p_0) + w$. Such a map linear map $L$ can then be shown to be unique, and thus may be referred to as the differential of $F$ at $p$. Reading about differentiability classes, it seems to me that higher differentiability is defined only in terms of partial derivatives, which would require coordinates.
With $v$ and $w$ defined as before, can't, say, the second derivative of $F$ at $p_0$ be defined as the unique bilinear map $L_2 : V \times V \to W$ satisfying$$\lim_{p \to p_0} \frac{\lVert w - L_1 v - \frac{1}{2} L_2[v,v] \rVert}{\lVert v \rVert} = 0$$where $L_1$ is the differential of $F$ at $p_0$? And how about generalizing to arbitrary smoothness of class $C^k$, defining it locally for a point $p_0$ via for each $i = 1, \dots k$ the existence of an $i$-linear map $L_i : V \times \cdots \times V \to W$ such that$$\lim_{p \to p_0} \frac{\lVert v - P_k v \rVert}{\lVert v \rVert},$$where$$P_k v = \sum_{i=1}^k \frac{L_i[v, \dots, v]}{i!}?$$Is this sensible? I'm doing a writeup on differential geometry covering some background material, and want to keep it as coordinate-agnostic as possible. Any insight would be greatly appreciated.