Let $(A,m_1)$ be a differential graded algebra, $m_2:A\otimes A\to A$ its product and $H^*A$ its homology, on which there is another product induced by $m_2$ which I'll call equally. I'm trying to follow the steps in page 9 of these notes, right before section 3.5 and I'm stuck right at the begining.
Let us choose a morphism of chain complexes $f_1:H^*A\to A$ inducing the identity. Since it is a morphism of chain complexes, each component should be a morphism of dg algebras, but since after this there is a failure of $f_1$ to commute with the product, I guess the author only means that it has to commute with the differentials. I think the most obvious way to choose it is just choosing representatives, which is clearly commutes with the differentials. The text says that there exists a degree $-1$ map $f_2:H^*A^{\otimes 2}\to A$ such that
$$f_1 m_2 = m_2 (f_1 ⊗ f_1) + m_1 f_2.$$
I don't see why this is true. Let's take the map which chooses representatives. Then
$$f_1m_2([a],[b])=f_1([a][b])=f_1([ab])=ab+m_1(c_1) \text{ for some } c_1\in A.$$
On the other hand
$$m_2(f_1\otimes f_1)(a,b)=(a+m_1(c_2))(b+m_1(c_3))=ab+am_1(c_3)+m_1(c_2)b+m_1(c_2)m_1(c_3).$$
Therefore, the difference between these operations is
$$m_1(c_1)-am_1(c_3)+m_1(c_2)b+m_1(c_2)m_1(c_3).$$
How can this expression take the form of $m_1f_2$ for a map $f_2:H^*A^{\otimes 2}\to A$ of degree $-1$?
First note that a morphism of chain complexes is just required to commute with the differentials in any case (and I don't see what it would mean that each component would be a morphism of dg algebras, since components are just vector spaces).
In this case, as you said $f_1: H^*(A)\to A$ is just required to (be linear and) pick for each $x\in H^n(A)$ a representative $f_1(x)\in A^n$. Then you want to show that there is $f_2: H^*(A)\otimes H^*(A)\to A$ such that $$ f_1(xy) = f_1(x)f_1(y) + m_1(f_2(x\otimes y)). $$
You showed that, by definition of the product on $H^*(A)$, there is some $c\in A$ such that $f_1(xy)=f_1(x)f_1(y)+m_1(c)$. Now you just have to see that $c$ can be chosen linearly as $c=f_2(x\otimes y)$.
But you can for instance choose a basis $(e_i)$ of $H^*(A)$ and define $f_2(e_i\otimes e_j)$ to be any valid $c$, and extend $f_2$ by linearity to the whole space. You can check that this gives a correct answer to the problem.