Degree of a map S^1 -> S^1

Let f,g : S^1 -> S^1 be continuous maps. When composed with the
projection p:[0,1] -> S^1, we obtain maps which can be lifted to [0,1]
-> R. Let these be denoted by f' and g', respectively. We can associate
to both f and g an integer, called the degree, given by f'(1) - f'(0)
and g'(1) - g'(0), respectively, which is independent of the lifting.
Now, let h = f o g, the composition of f and g.

In a previous post, I asked how I might prove that
deg(h) = deg(f) * deg(g), where * denotes integer multiplication. A
very neat solution to this exercise was suggested (by Rufus), in which
we simply use homotopy class representatives z |-> z^n (z \in C, |z| =
1), for which the fact is obvious (and then using standard results
about homotopy equivalances). However, I am still interested in the
following two questions, arising from the direct proof I initially
conceived:

(i) How can one construct an explicit formula for (f o g)' ?
(ii) Is it true that f(e(x)) = e(deg(f)*x), or more generally
e(deg(f)*g'(x)) = f(e(g'(x))) ?
(iii) If so, how can we prove (ii)?


*Possible counter-example to (ii): f: (x,y) |-> (x,-y) is homotopic to
the identity?
Thank you.

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