The statement is not true (in topological spaces or simplicial sets). The composition $f'' \cdot f'$ will only be multiplication by $d$ "up to higher Atiyah--Hirzebruch filtration".

For an explicit counterexample one can calculate the effect for the universal cover of $B\mathbb{Z}/2$ in zeroth stable cohomotopy. By the Segal conjecture $\pi^0(\Sigma^\infty B\mathbb{Z}/2_+)$ is the completion of the Burnside ring $A(\mathbb{Z}/2)$ at the augmentation ideal $I$. As an abelian group this is free with basis 1, the singleton $\mathbb{Z}/2$-set, and $T$, the free transitive $\mathbb{Z}/2$-set. Taking bases 1 and $H := T-2$, we have $H^2 = (T-2)^2 = T^2-4T+4 = 4-2T = -2H$, so the $I$-adic filtration agrees with the 2-adic filtration on $\mathbb{Z}\{H\}$. Thus we have
$$\pi^0(\Sigma^\infty B\mathbb{Z}/2_+) = \mathbb{Z}\{1\} \oplus \mathbb{Z}_2 \{H\},$$
where $\mathbb{Z}_2$ denotes the 2-adic integers.

The Becker--Gottlieb transfer map $\mathrm{trf} : \Sigma^\infty B\mathbb{Z}/2_+ \to \Sigma^\infty E\mathbb{Z}/2_+ = \mathbb{S}^0$ (associated to the covering map $\pi : E\mathbb{Z}/2 \to B\mathbb{Z}/2$) on cohomotopy sends the unit to

$$T = H+2 \in \pi^0(\Sigma^\infty B\mathbb{Z}/2_+),$$

so $\pi \circ \mathrm{trf} : \Sigma^\infty B\mathbb{Z}/2_+ \to \Sigma^\infty B\mathbb{Z}/2_+$ on zeroth cohomotopy sends a $\mathbb{Z}/2$-set $X$ to $\vert X \vert \cdot T$. This has twice the cardinality of $X$, but is not equal to $2X$ (for example if $X=1$).

(Note that $H$ restricts to zero on the 0-skeleton of $\Sigma^\infty B\mathbb{Z}/2_+$, so has positive Atiyah--Hirzebruch filtration: thus $T=2$ module higher Atiyah--Hirzebruch filtration, as I mentioned above.)