What does least fix point and greatest fix point mean in Safety games
Probably the most common fixpoint expressions in model checking are things like $\mu X.A\cup(B\cap\circ X)$ and $\nu X.A\cap(B\cup\circ X)$, where $\circ$ is some flavour of "next state" operator. That is, the least $X$ such that $X = A\cup(B\cap\circ X)$, and the greatest $X$ such that $X = A\cap(B\cup\circ X)$, respectively. More generally, we are talking about fixpoints of monotonic set transformers, i.e. functions $P({\cal S})\to P({\cal S})$ operating on the subsets of a given set $\cal S$. 

An intuitive way of thinking about least fixpoints is that they represent sets constructed bottom-up using constructors. Suppose you wanted to define the natural numbers as a subset of the reals: $n$ is a natural number iff it is $0$, or the successor of another natural number. This means that $\mathbb{N}$ is a fixpoint of the set transformer $F:S\mapsto\{0\}\cup succ(S)$. This is not enough, though: the set of all real numbers is also such a fixpoint. $\mathbb{N}$ is the least fixpoint, though, meaning that it only contains those elements which it has to in order to be a fixpoint. Computationally, this fixpoint is the limit you obtain by iterating the given set transformer $F$ on the empty set, gradually assembling the elements of $S$, i.e. the limit of $\emptyset,F(\emptyset),F(F(\emptyset)),\dots$.

In the case of $S:=\mu X.A\cup(B\cap\circ X)$ this means that $S$ contains exactly those elements which can reach $A$, satisfying $B$ until they get there; not requiring the least fixpoint would mean that you could additionally have arbitrary $B$-cycles, infinite $B$-paths never satisfying $A$, etc in your set.

For greatest fixpoints, you have the dual situation: the set contains all elements which are not explicitly eliminated by the given conditions. For $S=\nu X.A\cap(B\cup\circ X)$, this includes any element which satisfies $A$, and either also satisfies $B$ or has a successor in $S$. We pick the greatest fixpoint because we want to include everything which doesn't violate the requirements. Otherwise we might end up with an excessively restrictive set (such as the least fixpoint, which is just $A\cap B$ in this case). Computationally, this is the limit you get by iterating the transformer on the full set $\cal{S}$ of states, gradually eliminating all which fail the specification, i.e. the limit of ${\cal S},F({\cal S}),F(F({\cal S})),\dots$.