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))
Some remarks should clarify the computation: In general, if two formulas φ:⇔φ0⊕x∧δ (with δ:⇔φ0⊕φ1) and ψ:⇔ψ0⊕x∧ρ (with ρ:⇔ψ0⊕ψ1) are given, then we have φ⊕ψ⇔((φ0⊕ψ0)⊕x∧(δ⊕ρ)) (see pages 134-136 of the BDD chapter). Hence, the ⊕ operation has to be recursively applied to the high- and low-parts of the given FDDs φ and ψ which is exactly what the Apply algorithm for BDD does if it would be applied to the FDDs. Hence, we can simply (mis-)use the BDD algorithm for Apply with ⊕ and the FDDs (even though the Apply algorithm is meant to work with BDDs) and it will correctly compute the result. However, we have to use the FDD elimination rule.
In particular, to compute Apply(⊕,(0⊕b^BDD2FDD(n0)),(0⊕b^1)), we can proceed as follows: We consider the "low-parts" 0 and 0, and the "high-parts" BDD2FDD(n0) and 1, so that Apply(⊕,(0⊕b∧BDD2FDD(n0)),(0⊕b∧1)) is the same as (0⊕0)⊕b∧Apply(⊕,BDD2FDD(n0),1).
In the second case, we have to compute Apply(⊕,(0⊕c∧1),1) which is reduced by the Apply algorithm to (0⊕1)⊕c∧1 which is same as 1⊕c∧1.
