The definition of a differentiable vector field on a manifold I have a question regarding the following section from M. Spivak's Calculus on Manifolds:

Let $M$ be a $k$-dimensional manifold in $\mathbb{R}^n$ . . .
. . .  Suppose that $A$ is an open set containing $M$, and $F$ is a differentiable vector field on $A$ such that $F(x) \in M_x$ for each $x \in M$. If $f:W \to \mathbb{R}^n$ is a coordinate system, there is a unique (differentiable) vector field $G$ on $W$ such that $f_*(G(a))=F(f(a))$ for each $a \in W$. We can also consider a function $F$ which merely assigns a vector $F(x) \in M_x$ for each $x \in M$; such a function is called a vector field on $M$. There is still a unique vector field $G$ on $W$ such that $f_*(G(a))=F(f(a))$ for $a \in W$; we define $F$ to be differentiable if $G$ is differentiable. Note that our definition does not depend on the coordinate system chosen: if $g:V \to \mathbb{R}^n$ and $g_*(H(b))=F(g(b))$ for all $b \in V$, then the component functions of $H(b)$ must equal the component functions of $G(f^{-1}(g(b)))$, so $H$ is differentiable if $G$ is.

I don't think the last statement is correct: Consider the unit circle $M=\{x^2+y^2=1\} \in \mathbb{R}^2$ and the coordinate functions $f:(-1,1) \to \mathbb{R}^2:x \mapsto(x,\sqrt{1-x^2}),g:(-1,1) \to \mathbb{R}^2:y \mapsto (\sqrt{1-y^2},y)$. Around the point $(1/\sqrt{2},1/\sqrt{2})$ I got the relation
$H(y)=G(f^{-1}(g(y))) \frac{-x}{\sqrt{1-x^2}},$
so that the component functions are not equal.
Am I right? If so, how should I make sense of the above section?
Thank you!
 A: Let $\varphi$ be the coordinate change $\varphi(a) = g^{-1}(f(a))$ for all $a\in W \cap f^{-1}[g[V]]$. Then we can write
\begin{align}
F(f(a)) &= f_\ast(G(a))\\
&= (g\circ \varphi)_\ast(G(a))\\
&= g_\ast\bigl(\varphi_\ast(G(a))\bigr),
\end{align}
and since $f(a) = g(\varphi(a))$, the condition $F(g(b)) = g_\ast(H(b))$ becomes $F(f(a)) = g_\ast(H(\varphi(a)))$. Since $g_\ast$ is injective by definition of a coordinate system, it follows that
$$H(\varphi(a)) = \varphi_\ast(G(a)).$$
Now using the definition/representation of $\varphi_\ast$ by the Jacobi matrix, we find for the components
$$H_i(\varphi(a)) = \sum_{j=1}^k \frac{\partial\varphi_i}{\partial x_j}(a)\cdot G_j(a).$$
Thus the components are not equal (in general), but the components of $H$ are linear combinations with smooth coefficients of the components of $G$, and vice versa (since $\varphi_\ast$ is invertible), hence the components of the one are differentiable if and only if the components of the other are.
(Assuming we're talking of $C^\infty$-manifolds. For $C^r$-manifolds with a finite $r \in \mathbb{N}\setminus \{0\}$, we have only a well-defined notion of $C^s$-vector fields on $M$ for $s \leqslant r-1$, since we lose one order of differentiability by differentiating the coordinate change $\varphi$.)
