First consider the formula: $$\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt}$$
As we can see, $dx$ can be simplified from the RHS to get the LHS. This can be explained like this: define $y=x'(c)(t-c)+x(c)$ the tangent of $x(t)$ at $c$. Then $dx$ in $df/dx$ is simply $\Delta x$, while $dx$ in $\frac{dx}{dt}$ is $\Delta y$. And since $\Delta x\approx \Delta y$ when $\Delta t$ is very small, we can simplify $dx$ in the above formula.
Now consider the formula: $$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}$$
where $x,y$ are functions with variables $u,v$ and $f$ is function with variables $x,y$.
Of course we cannot simplify $\partial x$. I tried to use the same argument for $dx$, but it is more difficult to imagine. Maybe I need some exact definition of $\partial f$ and $\partial x$ (just a guess, maybe the definition is related to a tangent plane?)
Thanks so much.
Your tangent is now a tangent plane $$ f(x,y) = f(c,d) + \mbox{grad } f \cdot [(x,y) - (c,d)] $$ then $$ \Delta f = f(x,y) - f(c,d) = \mbox{grad } f \cdot (\Delta x, \Delta y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y $$