I would do that as given in the example solution in this case, i.e., first read the formulas form the BDD which is a&!c | !b&c, and the computing the Davio decomposition. Alternatively, you can construct a full DNF, which is

a&b&!c | a&!b&!c | a&!b&c | !a&!b&c

Then, you can replace OR by XOR

a&b&!c ⊕ a&!b&!c ⊕ a&!b&c ⊕ !a&!b&c

and then you replace the negations

a&b&(c⊕1) ⊕ a&(b⊕1)&(c⊕1) ⊕ a&(b⊕1)&c ⊕ (a⊕1)&(b⊕1)&c

multiply out

a&b&c ⊕ a&b ⊕ a&c ⊕ a ⊕ a&b&c ⊕ a&b ⊕ a&b&c ⊕ a&c ⊕ a&b&c ⊕ a&c ⊕ b&c ⊕ c

and clean up

a ⊕ a&c ⊕ b&c ⊕ c

which yields:

a ⊕ c&(a ⊕ 1 ⊕ b&1)

and that is the FDD. But maybe it is simpler to compute by cofactoring and directly computing the Davio composition as shown on the lecture slides.