Here is a simple example that confuses me regarding the orientation mappings:
In order to compute $\int_4^9{tdt}$ using $t=x^2$ parametrization in the interval $[-2,3]$ (i.e. $\int_{-2}^{3}{2x^3dx}$), one have to first prove that $t=x^2$ is an orientation preserving parametrization in $[-2,3]$. However, at $x=0$, the derivative of parametrization vanishes and orientation mapping inevitably jumps from +1 to -1.
In such a case, how can one prove that $t=x^2$ is an orientation preserving parametrization in intervals containing 0?
As you have correctly observed, $t=x^2$ is not an orientation preserving parameterization in any interval that contains $0$. The fact that the "change of coordinate" formula actually works in this example is a coincidence. Here's why.
If instead you were to use the interval $[2,3]$ (i.e. $\int_{2}^3 2x^3 \, dx$), then you could, of course, apply the change of parameter rules without trouble, to conclude that $$\int_4^9 t \, dt = \int_2^3 2x^3 \, dx $$ Here's the coincidence: $2x^3$ is an odd function, and therefore $\int_{-2}^2 2x^3 \, dx = 0$. It follows that $$\int_{-2}^3 2 x^3 \, dx = \underbrace{\int_{-2}^2 2 x^3 \, dx}_{=0} + \int_2^3 2x^3 \, dx = \int_2^3 2x^3 \, dx $$ If you tried to do this in a situation where the change-of-variables integrand was not an odd function, it would not work.