I managed to get an answer to my question however I should mention that the question itself has some missing relevant information that makes all the difference.
When trying to compute the second order correction to the anomalous magnetic moment of the electron what one usually tries to do is to express the transition current of the electron in the external magnetic field as
$$ \left
=\overline{u} (p')\left[\gamma^{\mu}F_1(q^2)+\frac{i\sigma^{\mu\nu}q_\nu}{2m}F_2(q^2)\right] u(p). $$ where $q = p'-p$ and $F_1$ and $F_2$ are the form factors. This is eq. 7, on page 196 of <PERSON>'s "Quantum Field Theory in a Nutshell" (2nd edition).
It is $F_2$(0) that gives us the anomalous magnetic moment.
One can use the <PERSON> decomposition identity,
$$ \overline{u} (p')\gamma^{\mu}u(p)=\overline{u} (p')\left[\frac{(p+p')^\mu}{2m}+\frac{i\sigma^{\mu\nu}(p'\nu-p\nu)}{2m}\right] u(p), $$
to write
$$ \left
=\overline{u} (p')\left[\gamma^{\mu}[F_1(q^2)+F_2(q^2)]-\frac{(p'+p)^\mu}{2m}F_2(q^2)\right] u(p). 818
$$
Usually, in the derivations found in the books, one tries to group all terms with $p^\mu$ and $p'^\mu$ to one side and those with $\gamma^\mu$ to the other. The anomalous magnetic moment will thus be the coefficient of the $(p'+p)^\mu$ term.
In the derivation that I was looking for I should have said that I was looking to prove that
$$ \overline{u}(p')[{p\mkern-7.5mu/}{\gamma_\mu} {p\mkern-7.5mu/'}]u(p) = \overline{u}(p')[2m(p_\mu + p'_\mu)]u(p) + \text{terms in $\gamma^\mu$}. $$
Now here is my proof. For it I will use the identity
$$ {a\mkern-7.5mu/}\gamma_\mu+\gamma_\mu{a\mkern-7.5mu/}=2a_\mu,\tag{1} $$
which can be obtained from the definition
$$ \gamma_\mu\gamma_\nu+\gamma_\nu\gamma_\mu=2g_{\mu\nu}, $$
doing the product with $a^\nu$ which is a generic four-vector (I will use it with $p$ and $p'$).
Starting from $\overline{u}(p')[{p\mkern-7.5mu/}{\gamma_\mu} {p\mkern-7.5mu/'}]u(p)$ and using (1) to replace ${p\mkern-7.5mu/}\gamma_\mu$ leads to
$$ \overline{u}(p'){p\mkern-7.5mu/}{\gamma_\mu}{p\mkern-7.5mu/'}u(p) = 2p^\mu\overline{u}(p'){p\mkern-7.5mu/'}u(p) -\overline{u}(p')\gamma_\mu{p\mkern-7.5mu/}{p\mkern-7.5mu/'}u(p).\tag{2} $$
Now I am going to use $$ {a\mkern-7.5mu/}{b\mkern-7.5mu/}=-{b\mkern-7.5mu/}{a\mkern-7.5mu/}+2(ab)\tag{3} $$ which is obtained from (1) by doing the product with $b^\mu$. 818