Let
$$R(\theta):=\begin{bmatrix}\cos\frac\theta2 & -\sin\frac\theta2 \\ \sin\frac\theta2 & \cos\frac\theta2
\end{bmatrix},\ \theta\in \mathbf R, \quad D := \begin{bmatrix}1 & 0 \\ 0 & -1
\end{bmatrix}; \\
S(\theta):=R(\theta)DR(\theta)^T=\begin{bmatrix}\cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix}, \quad \text{(diagonalization of $S(\theta)$)} \\
v := [|+\rangle\;|-\rangle],\; v^\dagger = \begin{bmatrix}\langle+|\\ \langle -| \end{bmatrix}, \\
C:=\begin{bmatrix}0&1 \\-1&0\end{bmatrix}, \\
|\Psi\rangle :=\frac1{\sqrt2}[|+\rangle\; |-\rangle]_AC \otimes\begin{bmatrix}|+\rangle\\ |-\rangle \end{bmatrix}_B; \\
u(\theta) := vR(\theta)=[u_+\;u_-].
$$
In the all the prior expressions, the subscript of a matrix applies to each entry of the matrix. I suppose $$\hat A(\theta_a)= v_A S(\theta_a)v^\dagger_A=u_ADu_A^\dagger=u^A_+u^{A\dagger}_+-u^A_-u^{A\dagger}_-, \\
\hat B(\theta_b)= v_B S(\theta_b)v^\dagger_B=u_BDu_B^\dagger=u^B_+u^{B\dagger}_+-u^B_-u^{B\dagger}_-; \;\text{(spectral decomposition or diagonalization of $\hat A$ and $\hat B$)}\\
\hat A(\theta_a)\otimes \hat B(\theta_b) =
(u^A_+\otimes u^B_+)(u^{A\dagger}_+\otimes u^{B\dagger}_+)
-(u^A_-\otimes u^B_+)(u^{A\dagger}_-\otimes u^{B\dagger}_+)
-(u^A_+\otimes u^B_-)(u^{A\dagger}_+\otimes u^{B\dagger}_-)
+(u^A_-\otimes u^B_-)(u^{A\dagger}_-\otimes u^{B\dagger}_-).
$$
With these,
\begin{align}
\langle \hat A(a)\hat B(b)\rangle &= \langle\Psi|\hat A(\theta_a)\otimes \hat B(\theta_b)|\Psi\rangle \\
&= |\langle\Psi|u^A_+\otimes u^B_+\rangle|^2-|\langle\Psi|u^A_-\otimes u^B_+\rangle|^2-|\langle\Psi|u^A_+\otimes u^B_-\rangle|^2+|\langle\Psi|u^A_-\otimes u^B_-\rangle|^2 \\
&= \mathcal P_{++}-\mathcal P_{-+}-\mathcal P_{+-}+\mathcal P_{--}.
\end{align}
which is the desired result.
However, if our sole aim is to prove the second equality of Equation (F-7), i.e.
$$\langle \hat A(a)\hat B(b)\rangle = -\cos\theta_{ab},$$
it is much more straightforward to compute directly. We show the following two approaches.
2.1 \begin{align}
&2\langle\Psi|\hat A(\theta_a)\otimes \hat B(\theta_b)|\Psi\rangle \\
=& (\langle +-|-\langle-+|)\hat A\otimes\hat B (|+-\rangle-|-+\rangle) \\
=& \langle +|\hat A|+\rangle\langle-|\hat B|-\rangle
-\langle-|\hat A|+\rangle\langle+|\hat B|-\rangle
-\langle +|\hat A|-\rangle\langle-|\hat B|+\rangle
+\langle-|\hat A|-\rangle\langle+|\hat B|+\rangle \\
=&A_{1,1}B_{2,2}-A_{2,1}B_{1,2}-A_{1,2}B_{2,1}+A_{2,2}B_{1,1} \qquad\qquad (A=S(\theta_a),\;B=S(\theta_b))\\
=&-\cos(\theta_a)\cos(\theta_b)-\sin(\theta_a)\sin(\theta_b)-\sin(\theta_a)\sin(\theta_b) -\cos(\theta_a)\cos(\theta_b) \\
=&-2\cos(\theta_a-\theta_b)
\end{align}
2.2 In general
\begin{align}
\langle\Psi|\hat A(\theta_a)\otimes \hat B(\theta_b)|\Psi\rangle &= \sum_{i,j,m,n}c_{i,j}^*(v_i^\dagger\otimes v_j^\dagger)\hat A\otimes\hat B\, c_{m,n}(v_m\otimes v_n) \\
&=\sum_{i,j,m,n}c_{i,j}^*\langle v_j,\hat Bv_n\rangle c_{m,n}\langle v_i,\hat Av_m\rangle \\
&=\sum_{i,j,m,n}c_{i,j}^*B_{j,n}c_{m,n}A_{i,m} \\
&=\frac12\operatorname{tr}(CBC^\dagger A^\dagger)
\end{align}
where $c_{i,j}$ is an entry of matrix $C$, $A$ and $B$ in the last two lines denotes matrices with respective entries $A_{i,m}=\langle v_i,\hat Av_m\rangle$ and $B_{j,n}=\langle v_j,\hat Bv_n\rangle$.
Here we have
$$A = S(\theta_a),\quad B=S(\theta_b),$$
and
$$CBC^\dagger A^\dagger=-\begin{bmatrix} \cos & -\sin \\ \sin & \cos\end{bmatrix}(\theta_a-\theta_b).$$
So
$$\langle\Psi|\hat A(\theta_a)\otimes \hat B(\theta_b)|\Psi\rangle = -\cos(\theta_a-\theta_b).$$
