Why does it work?

Now that we know how to use the chain, rule, let's see why it works. First recall the definition of derivative: $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x},$$ where $\Delta f = f(x+h)-f(x)$ is the change in $f(x)$ (the rise) and $\Delta x=h$ is the change in $x$ (the run).

From change in $x$ to change in $y$

In other words, whenever $\Delta x$ is small, $\frac{\Delta f}{\Delta x}$ is close to $f'(x)$. Another way of saying that is: $$ \hbox{if } \Delta x \approx0, \,\, \text{ then }\, \Delta f \approx f'(x) \Delta x.$$ Multiplying by $f'(x)$ converts changes in $x$ into changes in $f(x)$. We say that $f'(x)$ is a conversion factor from changes in $x$ to changes in $f(x)$.

From change in $x$ to change in $y$, passing through $u$

Now suppose that $y=f(u)$ and that $u=g(x)$. Then $\frac{dy}{du}=f'(u)$ is the conversion factor between changes in $u$ and changes in $y$. That is, $$\Delta y \approx f'(u) \Delta u.$$ Meanwhile, $\frac{du}{dx}=g'(x)$ is the conversion factor between changes in $x$ and changes in $u$. That is, $$\Delta u \approx g'(x) \Delta x.$$

Put those two results together to get $$\Delta y \approx f'(u) \Delta u \approx f'(u)g'(x) \Delta x,$$ which means that the conversion factor from $\Delta x$ to $\Delta y$ is $f'(u)g'(x)$, or equivalently that $$\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}.$$ That's the chain rule!