As suggested in the comments the diagram evaluates to $0$, introducing a photon mass $\mu$
\begin{equation}
\begin{split}
\text{Fourier amputated diagram} &= (-ie)^2(-1)\int \frac{d^4k}{2\pi^4} \gamma^{\mu} \text{Tr} \left[ \gamma^{\nu} \frac{\require{cancel}\cancel{k}+m}{k^2-m^2} \right] \frac{-i\eta_{\mu\nu}}{-\mu^2} \\
& \propto \int d^4k \, \gamma^\mu k_\alpha \frac{\text{Tr}\left[\gamma^\mu \gamma^\alpha \right]}{k^2-m^2} \\
& \propto \int d^4k \, \frac{\cancel{k}}{k^2-m^2} = 0
\end{split}
\end{equation}
You are right in pointing out the book's logic [Steps 1 & 2 in your post]. You are also right in deducing that the form factor $F_2(q^2)$ does not contribute to the infrared divergences of the loop corrected $\textbf{p}\rightarrow\textbf{p}'$ graphs and hence, the authors of P&S discard it for the study of the infrared divergent part of the loop corrections.
Remember that what we wish to study here is the infrared behavior of loop corrections in amplitudes that correspond to a fermion, scatterred off by some external potential. The tree level graph for that is simple. Adding virtual particles to the aforementioned graph can occur in two ways:
- Adding a virtual fermion: such graphs do not produce infrared divergences
- Adding a virtual photon: if the photon is SOFT (i.e. low energetic), such graphs correspond to infrared divergences, and hence, those graphs are the graphs we wish to study.
In the latter case (of the additional virtual photon being soft), the effect of adding the additional soft virtual photon in any tree level graph, corresponds to multiplying the amplitude corresponding to that tree level graph by a factor. This factor is precisely the for factor $F_1(q^2)$. $F_1(q^2)$ is the leading term in an expansion with respect to the soft momentum, carried by the additional virtual photon. The effect of adding virtual photons that are not soft, or the effects coming from adding the remaining terms of the soft expansion (that do not yield any infrared divergences), are accounted for in corrections to the tree-level amplitude that do not factor out, like $F_2(q^2)$. I suggest reading the paper of Weinberg, called "Infrared Photons and Gravitons", or Chap.13 from vol. I of its book on QFT.
Apart from the vertex corrections, there also exist the fermion self-energy graphs that may occur by adding a virtual photon onto an external fermion. Those also contain infrared divergences, but they are not considered here, as their effect is completely cancelled by renormalization (see Weinberg's vol.I in QFT, chapter 13.2 for more and see also a question I had asked about infrared divergences and self-energy graphs: Renormalization and virtual soft divergences).
Now that we have studied all the possible ways in which an infrared virtual divergence can occur, we must also study the emission of a single photon from a tree level $\textbf{p}\rightarrow\textbf{p}'$ graph. The amplitudes corresponding to that emission are also infrared divergent and we observe that the cross section corresponding to that amplitude exactly cancels the infrared divergences associated with the addition of a virtual soft photon into the tree level $\textbf{p}\rightarrow\textbf{p}'$ graph. The remaining (finite) part of the sum of the two cases is what we observe in experiments due to the finite resolution of the detectors (the detectors can not observe infinitely small frequencies/energies of the emitted photons).
So, to sum up, you are right in pointing out that $F_2(q^2)$ does not contain any infrared divergences, and is, therefore, beyond the scope of the analysis of sub-chapter 6.4 in P&S. I have just provided a more spherical, in my view, picture of what is going on here. I hope it helps. If anything is unclear, please comment.
Best Answer
The infrared (IR) divergencies comes from internal massless propagators, i.e. photon propagators. An infrared regulator can be introduced e.g. by giving the photon a mass.
Now let's turn to ultraviolet (UV) divergencies. It's difficult to give an ironclad/watertight argument for analyticity of correlator functions. In this answer we will just try to amend the explanation given in Ref. 1.
The electron self-energy $\Sigma(p,m)$ in $d=4$ has superficial degree of divergence (SDOD) $D=1$. Each time we differentiate wrt. the fermion momentum $p^{\mu}$ or the fermion mass $m$, we effectively gain 1 more fermion propagator, and hence lower the SDOD by 1 unit, cf. Ref. 1.
Lorentz covariance dictates that the tensor structure of $\Sigma$ comes from $\not p$.
We can therefore Taylor expand around $p=0=m$: $$ \Sigma(p,m)~=~ A_0\Lambda + \{a_0m + a_1 \not p\} \ln\Lambda + \text{finite terms}, \tag{10.6} $$ where $\Lambda$ is a UV momentum cut-off. The Taylor coeffcients are independent of $p$ and $m$ by definition, cf. OP's question.
If $m=0$, then the QED Lagrangian has chiral symmetry. Then the 1PI effective action (which contains $\Sigma$ at quadratic order) should also possess chiral symmetry, cf. Ref. 2. The $A_0$ term breaks chiral symmetry, and must hence be absent. Since it does not depend on $m$, it must also be absent for $m\neq 0$. Hence $\Sigma$ is actually only logarithmically divergent.
For a similar analysis of the divergent parts of the photon self-energy, see e.g. this Phys.SE post.
References:
M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 10.1, p. 319.
S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; Section 16.4.