The confusion is caused by using $R$ to denote two different things.
In Section 13.6, Besse introduces $R$ as a $6$-by-$6$ matrix of (scalar) curvature coefficients, which is the matrix of the linear transformation $\Lambda^2\to\Lambda^2$ induced by the curvature operator with respect to a basis adapted to the standard self-dual-and-anti-self-dual splitting of $\Lambda^2$. When you square this $R$ (which is the same as $R\wedge R$ since the entries are scalars), you will get a symmetric $6$-by-$6$ matrix whose trace is quadratic in the coefficients of the curvature matrix (and, in fact, it will be a positive definite quadratic form on the space of curvature tensors, not what you want at all). Note that you could see that the entries of $A$ and $D$ couldn't be $2$-forms because, if they were, it wouldn't make any sense to subtract $s/12$, which is clearly a scalar. (Actually, it doesn't make any sense anyway because $A$ and $D$ are $3$-by-$3$ matrices, but, never mind. This was only meant to be a heuristic indication of the real formulae anyway.)
Meanwhile, in that passage of Section 13.8, Besse is instead using $R$ to denote a skew-symmetric $4$-by-$4$ matrix of $2$-forms, or, more invariantly, a $2$-form with values in the skew-symmetric endomorphisms of the tangent bundle of $M$. (The entries of $A$, $B$, $C$, and $D$ appears as coefficients of the $2$-form entries in this $R$.) At this point, $R\wedge R$ is a symmetric $4$-by-$4$ matrix of $4$-forms, and the trace is a $4$-form on $M$. As for the appearance of the factors of $2$, that is wrapped up with Besse's choice of the norms on $W_\pm$ as spaces of linear transformations $\Lambda^2_\pm\to\Lambda^2_\pm$. These norms are unique up to a choice of a constant, but one has to make that choice consistently in order for the formulae to work out correctly.
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.