The space-time supersymmetry on left-handed modes is generated by
$$
q_{\alpha}^{q}=\oint\frac{dz}{2\pi i} \Sigma_{\alpha}^{(q)}(z),\qquad q\in \frac{\mathbb{Z}}{2}
$$
where the $\Sigma_{\alpha}^{(q)}(z)$ is the $q$-picture vertex operator of the Ramond ground state. This charge generate a symmetry that act on the physical states of the theory. However, there are an infinite tower of redundancies due to picture changing. The supersymmetry generator have half-integer picture.
A given property of a state, e.g. the fact that is annihilated by half the super-symmetries, does not depend on the picture you are using to represent it. The picture changing phenomena can be viewd as a residual gauge symmetry that comes from the gauge fixing of the superconformal world-sheet symmetry. The only thing that is fixed by the theory is the total picture of a given Riemann surface.
This is analogous to the case in bosonic string where the gauge fixing of the conformal world-sheet symmetry leads to a residual redundancy on which operator should be integrated and which should be fixed with $c\tilde{c}$ insertion. Since the $c$ ghost is nilpotent, we have a more restricted redundancy than the ones related with the superconformal ghost $\gamma$.
The answer for your first question is yes. The only novelty of this "current algebra" is that the picture number is not conserved by the algebra, so you should work with all pictures. At the end you can always doing a picture change back to satisfy a given prescription.
Note that this only works if the states are on-shell, i.e. if the states are BRST invariant. For off-shell states the problem of picture changing is much more involved.
Related to your second question the GSO projection $(-1)^{F}$ measure the chirality of the Ramond ground state.