Let $ \pi : P \rightarrow B$ and $\pi' ; Q \rightarrow B$ be two principal G- bundles.
Why this is true:
If $f : P \rightarrow Q $ is a morphism of the pricipal G-bundles P and Q ( i.e. $f(p.g)= f(p).g, \forall p \in P, \forall g \in G $), then $ \pi'\circ f = \pi $ ?
This isn't true as stated. For an easy counter example, let $P=Q=B \times G$, the trivial bundle. Then, given any nonidentity map $f:B \to B$ you get a map $P \to Q$ defined by $(x,g) \mapsto (f(x),g)$. This is certainly equivariant, but it doesn't satisfy the property you ask for.
There is a condition missing from the source you cite. Indeed, the answer at the linked page says that a morphism is an equivariant fiber bundle map, this means that it maps over the identity (i.e. satisfies the condition you ask for). The notes you mention just forgot to say this, its implicit in everything happening over the base space.