Assume $f$ is a smooth function, $(\phi, U)$ and $(\psi,U)$ are two coordinate system, $\phi(p)=\left(x_1(p),\ldots,x_n(p)\right)$, $\psi(p)=\left(y_1(p),\ldots, y_n(p)\right)$ and $$ y_i(p)=x_i(p)-x_i(m), $$ where $m\in U$ is a fix point.
I need to proof $$\frac{\partial f}{\partial x_i}(m)= \frac{\partial f}{\partial y_i}(m),$$
where $\frac{\partial f}{\partial x_i}(m)=\left.\frac{\partial}{\partial x_i}\right|_m f=\frac{\partial f\circ \phi^{-1}}{\partial u_i}(\phi(m))$, $\frac{\partial f}{\partial y_i}(m)=\left.\frac{\partial}{\partial y_i}\right|_m f=\frac{\partial f\circ \psi^{-1}}{\partial u_i}(\psi(m))$ and $u_i\circ(x_1,\ldots,x_n)=x_i$.
The $\phi^{-1}$ and $\psi^{-1}$, I do not know how to deal with them.
Before I get into how to formalize this, notice that what's going on here is that $\phi$ and $\psi$ differ by a translation, whose differential in matrix form is the identity matrix, so this really follows by the chain rule.
To write it out in coordinates, notice from the definition that if $T_z$ represents translation by fixed $z=(z_1,\cdots z_n)$, then
$$\frac{\partial u_j\circ T_z}{\partial u_i} = \frac{\partial (u_j+z_j)}{\partial u_i}=\delta^j_i\text{.}$$
Moreover, you have
\begin{align*} \psi(p) &= \phi(p)-\phi(m) =(T_{-\phi(m)}\circ \phi)(p)\\ \psi & = T_{-\phi(m)}\circ \phi\\ \psi^{-1} &= \phi^{-1}\circ T_{-\phi(m)}^{-1} = \phi^{-1}\circ T_{\phi(m)}\\ f\circ \psi^{-1} &= f\circ \phi^{-1}\circ T_{\phi(m)}\text{.} \end{align*}
And finally observe $\psi(m)=0$.
So now you can just use the chain rule in its coordinate form.
\begin{align*} \frac{\partial f\circ\psi^{-1}}{\partial u_i}(\psi(m)) &=\frac{\partial f\circ\phi^{-1}\circ T_{\phi(m)}}{\partial u_i}(0) \\ &=\sum_{j=1}^{n}\frac{\partial f\circ\phi^{-1}}{\partial u_j}(T_{\phi(m)}(0))\cdot \frac{\partial u_j\circ T_{\phi(m)}}{\partial u_i}(0)\\ &=\sum_{j=1}^{n}\frac{\partial f\circ\phi^{-1}}{\partial u_j}(\phi(m))\cdot \delta^j_i \\ &=\frac{\partial f\circ\phi^{-1}}{\partial u_i}(\phi(m))\text{.} \end{align*}