I have the following situation:

I'm trying to prove last point: the metric is invariant to re-parameterization of $f$.
So, the left member can be rewritten in the following way:
$$a^2 \int^{1}_{0}{\delta((r_{1}\circ \gamma)\dot{\gamma})\delta((r_{2}\circ \gamma)\dot{\gamma})\frac{1}{r(t)}\text{d}t}+b^{2}\int^{1}_{0}{\delta(\Theta_{1} \circ \gamma)\delta(\Theta_{2} \circ \gamma)r(t) \text{d}t}$$
and i tried $\delta((r_1 \circ \gamma)\dot{\gamma})=\delta(r_1 \circ \gamma)\dot{\gamma}+(r_1 \circ \gamma)\delta(\dot{\gamma})$? I am thinking is not OK.
Any help it will be appreciated.
It's not always true. A necessary and sufficient on $r$ and $\gamma$ for the reparameterization to be true for all $r_1,\Theta_1,r_2,\Theta_2$ is that $r(t) \equiv r(\gamma(t))\dot{\gamma}(t)$.
I'll drop the $\delta$, since it just clouds the notation pointlessly.
$$\left\langle \left((r_1\circ \gamma)\dot{\gamma},\Theta_1\circ \gamma\right),\left((r_2\circ \gamma)\dot{\gamma},\Theta_2\circ\gamma\right)\right\rangle = a^2\int_0^1 r_1(\gamma(t))r_2(\gamma(t))\dot{\gamma}(t)^2\frac{1}{r(t)}dt+b^2\int_0^1 \Theta_1(\gamma(t))\Theta_2(\gamma(t))r(t)dt.$$ Let $t' = \gamma(t), dt' = \dot{\gamma}(t)dt$ so that we get $$a^2\int_0^1 r_1(t')r_2(t')\frac{\dot{\gamma}(\gamma^{-1}(t'))}{r(\gamma^{-1}(t'))}dt'+b^2\int_0^1 \Theta_1(t')\Theta_2(t')\frac{r(\gamma^{-1}(t'))}{\dot{\gamma}(\gamma^{-1}(t'))}dt',$$ while $$\left\langle\left(r_1,\Theta_1\right),\left(r_2,\Theta_2\right)\right\rangle = a^2\int_0^1 r_1(t')r_2(t')\frac{1}{r(t')}dt'+b^2\int_0^1 \Theta_1(t')\Theta_2(t')r(t')dt'.$$ Just note $\frac{\dot{\gamma}(\gamma^{-1}(t'))}{r(\gamma^{-1}(t'))} = \frac{1}{r(t')}$ if and only if $r(t) = r(\gamma(t))\dot{\gamma}(t)$.