Let $A=\{(x,y,z)\in S^2:z= 0\}$ and $f:A\rightarrow S^1$ with $(x,y,z)\mapsto(-y(1-z^2),x(1-z^2))$.
I am looking for a $\omega$ s.t. for we get $\int_{S^1}\omega=1$ and $\int_{A}F^*\omega=1$.
Which $\omega$ can I take?
I tried $\omega=xdx$ but that does not give $\int_{S^1}\omega=1$.
Notice that $f\big|_A \big( (x , y, 0) \big) = (-y, x)$, therefore $f \big| _A$ is clearly a diffeomorphism, so as soon as we find $\omega$ on $S^1$ such that $\int _{S^1} \omega = 1$ it will automatically follow that $\int _A f^* \omega = \int _{f(A)} \omega = \int _{S^1} \omega$ (without the need for you to require this).
Now endow $S^1$ with the form $\omega = \frac 1 {2\pi} (x \ \Bbb d y - y \ \Bbb d x)$. Notice that, using any of the techniques shown in the answers to your previous question (I am going to use Stokes' theorem with $S^1 = \partial D_1$), the integral of this form is
$$\int \limits _{\partial D_1} \omega = \frac 1 {2\pi} \int \limits _{D_1} \Bbb d (x \ \Bbb d y - y \ \Bbb d x) = \frac 1 {2\pi} \int \limits _{D_1} 2 \ \Bbb d x \wedge \Bbb d y = \frac 1 \pi \int \limits _{D_1} \Bbb d x \Bbb d y = \frac 1 \pi \text{ area } (D_1) = 1 ,$$
therefore it is a form as required. This form is not unique though, there are infinitely many that integrate to $1$.