The right part of $f$ being not relevant, we consider the diffeo $f:[0,1/2]\to[0,1]$, $f(x)=x+2^\alpha x^{1+\alpha}$, whose inverse map $g:[0,1]\to[1,1/2]$ is strictly increasing with unique fixed point $0$, where it has the expansion $$g(x)=x\big(1-2^\alpha x^{\alpha}+o(x^{\alpha})\big).$$ (Incidentally, note that there is a power series expansion valid for $0\le x< \frac12\alpha{(1+\alpha)^{-1-\frac1\alpha}}$, that may be of use for further refinements of the asymptotic: $$g(x)=x\sum_{k=0}^\infty\frac{(-1)^k}{k\alpha +1}{k\alpha +k \choose k}(2x)^{k\alpha}=x-2^\alpha x^{1+\alpha}+(\alpha+1)2^{2\alpha} x^{1+2\alpha}-\dots$$ and for any power of $g$ as well). Since $g$ is continuous, increasing, with unique fixed point $0$, the iteration $$\cases{ x_0 \in [0,1]\\\\x_{n+1}=g(x_n)}$$ converges monotonically to $0$. Thus we have, for $n\to\infty$ $$x_{n+1}=x_n\big(1-2^\alpha x_n^{\alpha}+o(x_n^{\alpha})\big) .$$ $$x_{n+1}^\alpha=x_n^\alpha\big(1-2^\alpha x_n^{\alpha}+o(x_n^{\alpha})\big)^\alpha=$$$$=x_n^\alpha\big(1-\alpha2^\alpha x_n^{\alpha}+o(x_n^{\alpha})\big)=x_n^\alpha-\alpha2^\alpha x_n^{2\alpha}+o(x_n^{2\alpha}). $$ $$\frac{x_{n+1}^\alpha}{x_{n}^\alpha}=1 +o(1).$$ $$\frac1{x_{n+1}^\alpha}-\frac1{x_{n}^\alpha}=\frac{x_{n+1}^\alpha-x_{n}^\alpha}{x_{n+1}^\alpha x_{n}^\alpha}=\frac{\alpha2^\alpha x_n^{2\alpha}+o(x_n^{2\alpha})}{x_{n+1}^\alpha x_{n}^\alpha}=\frac{\alpha2^\alpha x_n^{\alpha}\big(1+o(1)\big)}{x_{n+1}^\alpha }=$$$$=\alpha2^\alpha\frac{ x_n^{\alpha}}{x_{n+1}^\alpha }\big(1+o(1)\big)=\alpha2^\alpha +o(1).$$ Therefore summing from $1$ to $n$ $$\frac1{x_{n}^\alpha}=n\alpha2^\alpha +o(n), $$ and finally $$ x_{n} =\frac{\alpha^{-1/\alpha}}2\, n^{-1/\alpha} \, (1 +o(1)). $$

One can also improve the above asymptotic inserting it into an expansion for $$\frac1{x_{n+1}^\alpha}-\frac1{x_{n}^\alpha}.$$

The right part of $f$ being not relevant, we consider the diffeo $f:[0,1/2]\to[0,1]$, $f(x)=x+2^\alpha x^{1+\alpha}$, whose inverse map $g:[0,1]\to[1,1/2]$ is strictly increasing with unique fixed point $0$, where it has the expansion $$g(x)=x\big(1-2^\alpha x^{\alpha}+o(x^{\alpha})\big).$$ (Incidentally, note that there is a power series expansion valid for $0\le x< \frac12\alpha{(1+\alpha)^{-1-\frac1\alpha}}$, that may be of use for further refinements of the asymptotic: $$g(x)=x\sum_{k=0}^\infty\frac{(-1)^k}{k\alpha +1}{k\alpha +k \choose k}(2x)^{k\alpha}=x-2^\alpha x^{1+\alpha}+(\alpha+1)2^{2\alpha} x^{1+2\alpha}-\dots$$ and for any power of $g$ as well). Since $g$ is continuous, increasing, with unique fixed point $0$, the iteration $$\cases{ x_0 \in [0,1]\\\\x_{n+1}=g(x_n)}$$ converges monotonically to $0$. Thus we have, for $n\to\infty$ $$x_{n+1}=x_n\big(1-2^\alpha x_n^{\alpha}+o(x_n^{\alpha})\big) .$$ $$x_{n+1}^\alpha=x_n^\alpha\big(1-2^\alpha x_n^{\alpha}+o(x_n^{\alpha})\big)^\alpha=$$$$=x_n^\alpha\big(1-\alpha2^\alpha x_n^{\alpha}+o(x_n^{\alpha})\big)=x_n^\alpha-\alpha2^\alpha x_n^{2\alpha}+o(x_n^{2\alpha}). $$ $$\frac{x_{n+1}^\alpha}{x_{n}^\alpha}=1 +o(1).$$ $$\frac1{x_{n+1}^\alpha}-\frac1{x_{n}^\alpha}=\frac{x_{n+1}^\alpha-x_{n}^\alpha}{x_{n+1}^\alpha x_{n}^\alpha}=\frac{\alpha2^\alpha x_n^{2\alpha}+o(x_n^{2\alpha})}{x_{n+1}^\alpha x_{n}^\alpha}=\frac{\alpha2^\alpha x_n^{\alpha}\big(1+o(1)\big)}{x_{n+1}^\alpha }=$$$$=\alpha2^\alpha\frac{ x_n^{\alpha}}{x_{n+1}^\alpha }\big(1+o(1)\big)=\alpha2^\alpha +o(1).$$ Therefore summing from $1$ to $n$ $$\frac1{x_{n}^\alpha}=n\alpha2^\alpha +o(n), $$ and finally $$ x_{n} =\frac{\alpha^{-1/\alpha}}2\, n^{-1/\alpha} \, (1 +o(1)). $$

For instance, with $x_0=1$, $\alpha=1/3$ and $n=100$, Maple gives $x_n=0.000015\dots$, while $\frac{\alpha^{-1/\alpha}}2\, n^{-1/\alpha}=0.000013.$

One can also improve the above asymptotic inserting it into an expansion for $$\frac1{x_{n+1}^\alpha}-\frac1{x_{n}^\alpha}.$$