Translating intuition into rigor. The chain rule. 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
$$\frac{{df}}{{dx}} = \frac{{df}}{{dg}}\frac{{dg}}{{dx}}$$

Your intuition is solid, and the fact that $\Delta x \rightarrow 0 \implies \Delta g \rightarrow 0$ follows from the continuity of $g$.
There is a subtlety though: what if $\Delta g = 0$ for arbitrarily small $\Delta x$?  It is still possible to push the proof through, mostly just by thinking carefully about what this means.  I do this in $\S 5.2$ of these lecture notes.
