Added remark (at the request of the OP): The point is that, with respect to any local orthonormal coframing, the first Pontrjagin form has to be of the form $p_1 = Q(R)\omega$ where $Q(R)$ is quadratic in the entries of the $6$-by-$6$ matrix $R$ and $\omega$ is the local volume form associated to the coframing. Because of the way that $\mathrm{SO}(4)$ acts on $R$ under change of oriented coframe, $p_1$ must be of the form $$ p_1 = (c_1\ s^2 + c_2\ |B|^2 + c_3\ |W_+|^2 + c_4\ |W_-|^2)\ \omega $$ for some universal constants $c_i$ (since $R$ breaks up into 4 inequivalent representation of $\mathrm{SO}(4)$). Now, $p_1$ doesn't depend on a choice of orientation, nor does $s$ nor $B$, but moving to a different orientation will switch $W_+$ and $W_-$ and will also replace $\omega$ by $-\omega$. It follows that the only way the right hand side will be uninfluenced by the choice of orientation is if it is of the form $c\bigl(|W_+|^2-|W_-|^2\bigr)\ \omega$ for some universal constant $c$. The constant $c$ must be positive because $p_1(\mathbb{CP}^2)=3$, and $\mathbb{CP}^2$ has $W_- = 0$. (You can now determine that $8\pi^2c = 1$ by simply doing the integral on $\mathbb{CP}^2$, if you want.)

Added remark (at the request of the OP): The point is that, with respect to any local orthonormal coframing, the first Pontrjagin form has to be of the form $p_1 = Q(R)\omega$ where $Q(R)$ is quadratic in the entries of the $6$-by-$6$ matrix $R$ and $\omega$ is the local volume form associated to the coframing. Because of the way that $\mathrm{SO}(4)$ acts on $R$ under change of oriented coframe, $p_1$ must be of the form $$ p_1 = (c_1\ s^2 + c_2\ |B|^2 + c_3\ |W_+|^2 + c_4\ |W_-|^2)\ \omega $$ for some universal constants $c_i$ (since $R$ breaks up into 4 inequivalent representation of $\mathrm{SO}(4)$). Now, $p_1$ doesn't depend on a choice of orientation, nor does $s$ nor $B$, but moving to a different orientation will switch $W_+$ and $W_-$ and will also replace $\omega$ by $-\omega$. It follows that the only way the right hand side will be uninfluenced by the choice of orientation is if it is of the form $c\bigl(|W_+|^2-|W_-|^2\bigr)\ \omega$ for some universal constant $c$. The constant $c$ must be positive because $p_1(\mathbb{CP}^2)=3$, and $\mathbb{CP}^2$ has $W_- = 0$. (You can now determine that $4\pi^2c = 1$ by simply doing the integral on $\mathbb{CP}^2$, if you want. Of course, this depends on which norm you chose for $W_\pm$; I'm assuming you choose it to agree with Besse's choice. :) )

I think that one is expected to write out the translation between the two meanings and figure out which choice of norm on $W_\pm$ is meant when one is learning the subject.

Added remark (at the request of the OP): The point is that, with respect to any local orthonormal coframing, the first Pontrjagin form has to be of the form $p_1 = Q(R)\omega$ where $Q(R)$ is quadratic in the entries of the $6$-by-$6$ matrix $R$ and $\omega$ is the local volume form associated to the coframing. Because of the way that $\mathrm{SO}(4)$ acts on $R$ under change of oriented coframe, $p_1$ must be of the form $$ p_1 = (c_1\ s^2 + c_2\ |B|^2 + c_3\ |W_+|^2 + c_4\ |W_-|^2)\ \omega $$ for some universal constants $c_i$ (since $R$ breaks up into 4 inequivalent representation of $\mathrm{SO}(4)$). Now, $p_1$ doesn't depend on a choice of orientation, nor does $s$ nor $B$, but moving to a different orientation will switch $W_+$ and $W_-$ and will also replace $\omega$ by $-\omega$. It follows that the only way the right hand side will be uninfluenced by the choice of orientation is if it is of the form $c\bigl(|W_+|^2-|W_-|^2\bigr)\ \omega$ for some universal constant $c$. The constant $c$ must be positive because $p_1(\mathbb{CP}^2)=3$, and $\mathbb{CP}^2$ has $W_- = 0$. (You can now determine that $8\pi^2c = 1$ by simply doing the integral on $\mathbb{CP}^2$, if you want.)

I think that one is expected to write out the translation between the two meanings and figure out which choice of norm on $W_\pm$ is meant when one is learning the subject.

Remark: The notation in Besse's Sections 13.6 and 13.8 appears to be drawn, with little modification, from the famous 1978 paper "Self-Duality in Four-dimensional Riemannian geometry" by Atiyah, Hitchin, and Singer. However, the use of $\mathrm{tr}(R\wedge R)$ in the formula in Besse 13.8 seems to have been spliced in from the standard formula for the first Pontrjagin form; it is not in AHS where, instead, a different expression appears. I suspect that Besse also copied the choice of norm for the Weyl tensor from AHS, since they get the very same coefficient.