Given the function $f(x) = \begin{cases} 2^{x}-2, & x\leqslant 1 \\ \log _{2}(x-1), & x > 1 \end{cases}$,

First, find $f\left( \frac {5}{2}\right)$:
Since $\frac{5}{2} > 1$, use the second part of the function definition: $f\left( \frac {5}{2}\right) = \log _{2}\left( \frac {5}{2}-1\right) = \log _{2}\left( \frac {3}{2}\right)$.

Next, find $f\left[f\left( \frac {5}{2}\right)\right]$:
Substitute the result obtained above into the function: $f\left[\log _{2}\left( \frac {3}{2}\right)\right]$. Since $\log _{2}\left( \frac {3}{2}\right) \leqslant 1$, use the first part of the function definition: $f\left[\log _{2}\left( \frac {3}{2}\right)\right] = 2^{\log _{2}\left( \frac {3}{2}\right)} - 2 = \frac {3}{2} - 2 = -\frac {1}{2}$.

Hence, $f\left[f\left( \frac {5}{2}\right)\right] = \boxed{-\frac{1}{2}}$.