When considering two functions $f(x)$ and $g(x)$, it is known that
\left(f\circ g(x)\right)' = f'\circ g(x)\cdot g'(x)
So my intuitive approach is:
$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {g\left( {x + \Delta x} \right)} \right) - f\left( {g\left( x \right)} \right)}}{{\Delta x}}$
$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {g\left( {x + \Delta x} \right)} \right) - f\left( {g\left( x \right)} \right)}}{{g\left( {x + \Delta x} \right) - g\left( x \right)}}\frac{{g\left( {x + \Delta x} \right) - g\left( x \right)}}{{\Delta x}}$
Put $g\left( {x + \Delta x} \right) - g\left( x \right) = \Delta g\left( x \right)$
$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {g\left( x \right) + \Delta g\left( x \right)} \right) - f\left( {g\left( x \right)} \right)}}{{\Delta g\left( x \right)}}\frac{{g\left( {x + \Delta x} \right) - g\left( x \right)}}{{\Delta x}}$
So I guess the problem boils down to translating how $\Delta x \to 0 \Rightarrow \Delta g\left( x \right) \to 0$ and to adress ${\Delta g\left( x \right)}$'s behaviour.
The last intuition is to recklessly write
$g\left( {x + \Delta x} \right) - g\left( x \right) = \Delta g$
and put
$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta f\left( {g\left( x \right)} \right)}}{{\Delta g\left( x \right)}}\frac{{\Delta g\left( x \right)}}{{\Delta x}}$
which is the idea behind
$\frac{{df}}{{dx}} = \frac{{df}}{{dg}}\frac{{dg}}{{dx}}$