integral on a half circle

By definition, if the domain of $\gamma$ is $[a,b]$, then$$\int_\gamma\frac1{z^2}\,\mathrm dz=\int_a^b\frac{\gamma'(t)}{\gamma^2(t)}\,\mathrm dt$$and it seems to me that you are not taking that $\gamma'(t)$ into account.

To be more concrete, take $\gamma\colon[0,\pi]\longrightarrow\Bbb C$ defined by $\gamma(t)=e^{it}$ and take $\eta\colon[\pi,2\pi]\longrightarrow\Bbb C$ also defined by $\eta(t)=e^{it}$. Then\begin{align}\int_\eta\frac1{z^2}\,\mathrm dz&=\int_\pi^{2\pi}\frac{ie^{it}}{e^{2it}}\,\mathrm dt\\&=\int_0^\pi\frac{ie^{i(t+\pi)}}{e^{2i(t+\pi)}}\,\mathrm dt\\&=-\int_0^\pi\frac{ie^{it}}{e^{2it}}\,\mathrm dt,\end{align}since $e^{\pi i}=-1$ and $e^{2\pi i}=1$. But the final integral is just $\int_\gamma\frac1{z^2}\,\mathrm dz$.

The same argument applies if we have any even function instead of $z^2$.