The Poincare derivative has precisely the non-phase monodromy multipliers

Define $\mathrm{\Psi }(t,x)=\psi ({\varphi }_t(x))$. At $(T,x_{\ast })$, $\mathrm{\Psi }(T,x_{\ast })=0$ and ${\mathrm{\partial }}_t\mathrm{\Psi }(T,x_{\ast })=c^{\mathrm{\top }}f(x_{\ast })\mathrm{\ne }0$. By [F4], there is a local return-time function $\tau (x)$ on the section with $\tau (x_{\ast })=T$ and $\mathrm{\Psi }(\tau (x),x)=0$. Differentiating this identity at $x_{\ast }$ in a tangent direction $v\in S:=\ker c^{\mathrm{\top }}$ gives $D\tau (x_{\ast })v=-c^{\mathrm{\top }}Mv\mathrm{/}(c^{\mathrm{\top }}f(x_{\ast }))$.

The local return map is $P(x)={\varphi }_{\tau (x)}(x)$. The chain rule and step 1.1 give, for $v\in S$, $DP(x_{\ast })v=Mv-\frac{f(x_{\ast })c^{\mathrm{\top }}Mv}{c^{\mathrm{\top }}f(x_{\ast })}=\mathrm{\Pi }Mv,\mathrm{\Pi }=I-\frac{f(x_{\ast })c^{\mathrm{\top }}}{c^{\mathrm{\top }}f(x_{\ast })}\mathrm{.}$ The map $\mathrm{\Pi }$ identifies the quotient ${\mathbb{C}}^d\mathrm{/}\mathrm{span}\{f(x_{\ast })\}$ with $S$: it kills the phase direction and is the identity on $S$. Since $Mf(x_{\ast })=f(x_{\ast })$ by [F3], it induces a linear map on this quotient, and $DP(x_{\ast })$ is its conjugate under that identification.

In a basis whose first vector is $f(x_{\ast })$, the matrix of $M$ is block upper triangular with first diagonal block $1$ and remaining block the induced quotient map. Its characteristic polynomial is therefore $(\lambda -1)$ times that of the quotient map. The eigenvalue $1$ is simple by hypothesis, so the quotient eigenvalues, with multiplicity, are exactly all the remaining eigenvalues of $M$. Step 2.1 identifies these with the Poincare derivative eigenvalues.