Why can chain rule not be applied recursively (reapplied to outer function)? $f(x)=\sqrt{x+1}$  
$f(x)=(x+1)^{\frac 12}$  
$\frac {d}{dx}f(x)=\frac {d}{dx}[(x+1)^{\frac 12}]·\frac {d}{dx}(x+1)$  
$\frac {d}{dx}f(x)\neq[\frac {d}{dx}[(x+1)^{\frac 12}]·\frac {d}{dx} (x+1)]·\frac {d}{dx}(x+1)$  


Why can chain rule not be reapplied to $\frac {d}{dx}[(x+1)^{\frac 12}]$ in step three even though it is also a composite function (it looks the same as $f(x)$). Is it just a rule that chain rule can only be applied once? Does Leibniz's notation explain it?
Chain rule is the following:

$$ [f(g(x))]' = f'(g(x))g'(x) $$

So, when we want differentiate things like $\sqrt{1+x}$, we do $f(x) = x^{1/2}$ and $g(x)= 1+x$ . Then, we get

$$f'(x) = \frac{1}{2\sqrt{x}} \implies f'(g(x)) = \frac{1}{2\sqrt{1+x}}$$

$$g'(x) = 1 $$

Replacing in the formula for the chain rule:

$$ [f(g(x))]' = f'(g(x))g'(x)  $$
$$ (\sqrt{1+x})' =  \frac{1}{2\sqrt{1+x}} $$

Maybe you are confused with the notation.
The chain rule can be expressed as $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$. It logically follows that  $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dv} \frac{dv}{dx}$.