You have to consider the function BDD2FDD and also the Apply algorithm to understand the single steps. Indeed, the computation of the example solution has some unclarities which should be better explained here:

BDD2FDD(n3)
= BDD2FDD(n2) ⊕ a ∧ Apply(⊕, BDD2FDD(n1), BDD2FDD(n2))
// BDD2FDD(n2) = BDD2FDD(b) = (0⊕b∧1)
= (0⊕b∧1)⊕a∧Apply(⊕,BDD2FDD(n1),(0⊕b∧1))
// BDD2FDD(n1) = 0⊕b∧BDD2FDD(n0)
= (0⊕b∧1)⊕a∧Apply(⊕,(0⊕b∧BDD2FDD(n0)),(0⊕b∧1))
// Apply(⊕,(0⊕b∧BDD2FDD(n0)),(0⊕b∧1))

= (0⊕0)⊕b∧Apply(⊕,BDD2FDD(n0),1)
= (0⊕b∧1)⊕a∧(0⊕b∧Apply(⊕,(BDD2FDD(n0)),1))
// BDD2FDD(n0) = 0⊕c∧1
= (0⊕b∧1)⊕a∧(0⊕b∧Apply(⊕,(0⊕c∧1),1))
// Apply(⊕,(0⊕c∧1),1) = (0⊕1)⊕c∧1
= (0⊕b∧1)⊕a∧(0⊕b∧(1⊕c∧1))