Differential with Chain rule. (Relating to manifold.) Where did I make a mistake? Let $M$ be $n$-dimensional manifold, and $I\subset \mathbb R$ open interval.
For $p\in M$, define $L_p=\{ l : (-\epsilon, \epsilon)\to M \mid \epsilon >0, l(0)=p \}.$
Fix $p\in M$ and $l\in L_p.$
There exists $\epsilon>0$ s.t. $l:(-\epsilon, \epsilon)\to M$.
Let $(U, \phi)$ and $(V, \psi)$ be charts around $p$.
($\phi$ is a homeomorphism from $U$ to $\mathbb R^n$, and the same is true for $\psi.$)
Then, the coordinate transformation $\psi \circ \phi^{-1} : \phi (U\cap V) (\subset \mathbb R^n) \to \psi(U\cap V) (\subset \mathbb R^n)$, by $(x_1, \cdots , x_n)\mapsto (\psi \circ \phi^{-1})(x_1, \cdots, x_n)=:(y_1, \cdots, y_n)$ is homeomorphism.
Choose $\epsilon_0\in (0,\epsilon)$ s.t. $l((-\epsilon_0, \epsilon_0))\subset U$ and $l((-\epsilon_0, \epsilon_0))\subset V$ and assume the domain of $l$ is $(-\epsilon_0,\epsilon_0)$.
And let us write
$(\phi \circ l )(t)=(u_1(t), \cdots, u_n(t))$,
$(\psi \circ l) (t)=(v_1(t), \cdots, v_n(t))$.
Then, show that $$\dfrac{dv_k}{dt}=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} \dfrac{du_j}{dt}.$$

Here is my calculation and maybe I mistake somewhere.
Let $\psi \circ \phi^{-1}=:F, \phi \circ l=G.$
First, $\dfrac{d}{dt}(\psi \circ l)(t)=(\frac{d}{dt}v_1(t),\cdots, \frac{d}{dt}v_n(t))$.
On the other hand,
\begin{align}
&\quad \dfrac{d}{dt}(\psi \circ l)(t)\\
&=\dfrac{d}{dt}(F\circ G)(t)\\
&\underset{\mathrm{chain\ rule}}=(DF)(G(t))\cdot \dfrac{d}{dt}G(t)\ (DF:\mathrm{Jacobi\  matrix \ of\ } F)\\
&=\begin{pmatrix}
\frac{\partial y_1}{\partial x_1}(G(t)) & \cdots & \frac{\partial y_1}{\partial x_n}(G(t))\\
\vdots & & \vdots \\
\frac{\partial y_n}{\partial x_1}(G(t)) & \cdots & \frac{\partial y_n}{\partial x_n}(G(t))
\end{pmatrix}
\begin{pmatrix} \frac{d}{dt}u_1(t)\\ \vdots \\ \frac{d}{dt}u_n(t)\end{pmatrix}\\
&=\begin{pmatrix}
\displaystyle\sum_{j=1}^n \dfrac{\partial y_1}{\partial x_j}(G(t)) \dfrac{du_j}{dt}(t)\\
\vdots\\
\displaystyle\sum_{j=1}^n \dfrac{\partial y_n}{\partial x_j} (G(t))\dfrac{du_j}{dt}(t)
\end{pmatrix}.
\end{align}
Thus I get
$$\dfrac{dv_k}{dt}(t)=\displaystyle\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} (G(t))\dfrac{du_j}{dt}(t) \tag{i} $$
I think this is a bit different from the goal
$$\displaystyle\dfrac{dv_k}{dt}=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} \dfrac{du_j}{dt} \tag{ii}$$
Of course, the LHS of (i) :$\displaystyle\dfrac{dv_k}{dt}(t)$ is the same as the LHS of (ii) :$\dfrac{dv_k}{dt}$.
Similarly, $\dfrac{du_j}{dt}(t)=\dfrac{du_j}{dt}$.
But I don't think $\dfrac{\partial y_k}{\partial x_j} (G(t))=\dfrac{\partial y_k}{\partial x_j}$.
$\dfrac{\partial y_k}{\partial x_j}$ is written as $\dfrac{\partial y_k}{\partial x_j}(x_1, \cdots, x_n)$ but $\dfrac{\partial y_k}{\partial x_j} (G(t))
=\dfrac{\partial y_k}{\partial x_j} (u_1(t),\cdots, u_n(t)).$
So I think I mistook somewhere. Do you find where I mistook ?
 A: What you have done is absolutely correct. The problem is that the formula
$$\dfrac{dv_k}{dt}=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} \dfrac{du_j}{dt} \tag{1}$$
does not give you precise information at which points the derivatives $\dfrac{dv_k}{dt}$, $\dfrac{\partial y_k}{\partial x_j}$ and $\dfrac{du_j}{dt}$ have to be evaluated. It is clear that both $\dfrac{dv_k}{dt}$ and $\dfrac{du_j}{dt}$ have to be evaluted at the same $t$, but what about $\dfrac{\partial y_k}{\partial x_j}$? The partial derivatives $\dfrac{\partial y_k}{\partial x_j}$ are functions of the generic variable $\mathbf x = (x_1,\ldots,x_n)$, but it is not expedient to write
$$\dfrac{dv_k}{dt}(t)=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} (\mathbf x) \dfrac{du_j}{dt}(t) \tag{2}$$
because it does not tell us which specific $\mathbf x$ has to be taken. In fact we must take $\mathbf x = G(t)$:
$$\dfrac{dv_k}{dt}(t)=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} (G(t)) \dfrac{du_j}{dt}(t) \tag{3}$$
This is your formula $(i)$ and it is the only correct interpretation of $(1)$. It generalizes the chain rule from elementary calculus which says
$$(g \circ f)'(x) = g'(f(t)) f'(x)$$
or $$(g \circ f)' = (g' \circ f) \cdot f'$$
You would never write $(g \circ f)'(x) = g'(y) f'(x)$. But note that sometimes people write
$$\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} .$$
Here $y = y(x)$ and $z = z(y)$. This notation has the same problem as we encountered above.
A: It is a matter of definition: remember that when you have coordinates $\phi=(x_1,\dots,x_n):U\to\mathbb{R}^n$, $U\subset M$, and a function $f:M\to\mathbb{R}$, then $\frac{\partial f}{\partial x^i}$ is a function defined on $U$ by
$$\frac{\partial f}{\partial x^i}(p)=\frac{\partial (f\circ\phi^{-1})}{\partial r^i}(\phi(p)),$$
where $\frac{\partial}{\partial r^i}$ denotes the usual $i$-th partial derivative for functions from $\mathbb{R}^n$ to $\mathbb{R}$. Here, the function $F:\mathbb{R}^n\to\mathbb{R}^n$ you are trying to differentiate checks that
$$F(x)=\psi\circ\phi^{-1}(x)=(y_1\circ\phi^{-1}(x),\dots,y_n\circ \phi^{-1}(x)).$$
Thus, its Jacobi matrix at $x\in\mathbb{R}^n$ will be given by
$$\begin{pmatrix}
\frac{\partial (y_1\circ\phi^{-1})}{\partial r^1}(x)&\dots&\frac{\partial (y_1\circ\phi^{-1})}{\partial r^n}(x)\\
\vdots&&\vdots\\
\frac{\partial (y_n\circ\phi^{-1})}{\partial r^1}(x)&\dots&\frac{\partial (y_n\circ\phi^{-1})}{\partial r^n}(x)
\end{pmatrix}.$$
Then, by evaluating it at $G(t)=\phi(l(t))$ you get (for example
)
$$\frac{\partial (y_1\circ\phi^{-1})}{\partial r^n}(\phi(l(t)))=:\frac{\partial y_1}{\partial x^n}(l(t)),$$
and then proceeding as you did, you make explicit what was implicit in your given formula, which is intended to be read as:
$$\frac{dv_k}{dt}(t)=\sum_{j=1}^n\frac{\partial y_k}{\partial x^j}(l(t))\frac{du_j}{dt}(t)$$
or (evaluating at $t=0$)
$$\frac{dv_k}{dt}=\sum_{j=1}^n\frac{\partial y_k}{\partial x^j}(p)\frac{du_j}{dt}.$$
