The question really is about the difference between the susy variations of fields and their values in a given state.
Let's consider the transfomations in a $\mathcal N=1$ theory (very schematically). We have a chiral multiplet $\left\{\phi,\psi\right\}$ with variations
$$\delta_\epsilon\phi\sim \bar\epsilon\psi \,, \qquad\qquad \delta_\epsilon\psi\sim \partial_\mu \phi \gamma^\mu\epsilon\,,\tag{$*$}$$
and the vector multiplet $\left\{\lambda,A_\mu\right\}$ with variations
$$\delta_\epsilon \lambda\sim \partial_\mu A_\nu\gamma^{\mu\nu}\epsilon\,,\qquad\qquad \delta_\epsilon A_\mu \sim \bar\epsilon\gamma_\mu\lambda\,.\tag{$\dagger$}$$
In particular, in both cases we have
$$\delta\left(\text{boson}\right)\sim\text{parameter}\cdot\text{fermion} $$
and
$$ \delta\left(\text{fermion}\right)\sim\text{parameter}\cdot \partial\left(\text{boson}\right)\,.$$
(This is basically required by the susy algebra which implies $\left[\delta,\delta\right]\sim\partial$)
Now, when we look for a suspersymmetric configuration, we mean a set of field values such the state is invariant under susy variations. What fields can get VEVs?
- $\phi$ is a scalar, so its VEV respects Lorentz symmetry. It will break gauge symmetry if it's charged.
- $\psi$ and $\lambda$ are fermions, so a VEV would break Lorentz symmetry, and possibly gauge symmetry.
- $A_\mu$ is a vector, so it again breaks Lorentz symmetry and possibly gauge symmetry, but this is more subtle.
Phenomenologically, we know that Lorentz symmetry is unbroken, so usually, states with VEVs for fermions or vectors are not considered. Gauge symmetry may or may not be broken, as in QED vs. weak interactions. Hence, we are looking for a state with $\langle\psi\rangle=\langle\lambda\rangle=0$.
Finally, we explicitly want supersymmetric states, i.e. states that do not change under a susy transformation. In particular, since
$$0=\langle\psi\rangle\longrightarrow\langle\psi\rangle+\langle\delta_\epsilon\psi\rangle=\langle\delta_\epsilon\psi\rangle\,,$$
we require that the susy variations of $\psi$ and $\lambda$ vanish (in this state, i.e. with the chosen values for the other fields). Inspecting the variations $(*)$ and $(\dagger)$ above, this translates to $$\langle\partial_\mu\phi\rangle=0 \qquad \text{and} \qquad \langle F_{\mu\nu}\rangle=0\,.$$
This is what the paper means with
... any susy configuration must obey $F=0$ ...
