If $\omega$ and $\eta$ are closed differential forms on $M$ (compact of genus $g$), then
$$\int_M \omega\wedge \eta=\sum_{i=1}^g \int_{a_i}\omega\int_{b_i}\eta-\int_{a_i}\eta\int_{b_i}\omega $$
The core of the proof is to use the decompositions (we call them harmonic decompositions)
$$\omega=\sum_{i=1}^{2g}\mu_j\alpha_j$$
and
$$\eta=\sum_{i=1}^{2g}\nu_j\alpha_j,$$
denoting by $\alpha_j$ the basis dual to the canonical homology basis $\{n_j\}:=\{a_a,\dots,a_g,b_1,\dots,b_g\}$. In other words $\int_{a_k}\alpha_j=\delta_{kj}$ and
$\int_{b_k}\alpha_j=\delta_{rj}$, with $r=j-g$. This implies
$$\mu_j=\int_{n_j}\omega,$$
$$ \nu_j=\int_{n_j}\eta.$$
The existence of the dual basis is proven using the differential form associated to the cycles $a_i$ and $b_j$: this is a standard construction.
The harmonic decompositions for $\omega$ and $\eta$ are motivated by the following argument: the integral $\int_M \omega\wedge \eta$ is unchanged if we apply the replacement $\omega\rightarrow \omega+df$, wih $f$ $C^2$-function. Same holds for the r.h.s., as a direct computation shows.
As we can write any harmonic differential as the sum of a closed differential with an exact one (this results holds on $M$ compact), we arrive at the harmonic decompositions.
Using the harmonic decompositions, the proof of the integral formula follows once we compute the integrals of the form (called intersection numbers)
$$\int_{M}\alpha_j\wedge\alpha_k $$
which arise from the l.h.s. of the integral formula. The above intersection numbers are s.t.
$$\int_{M}\alpha_j\wedge\alpha_{j+g}=1 $$
for $j=1,\dots,g$ and
$$\int_{M}\alpha_j\wedge\alpha_{j-g}=-1 $$
for $j=g+1,\dots,2g$. This follows from the duality of the $\alpha_i$ with the canonical homology basis.
We are now ready to collect all results, i.e.
$$\int_M \omega\wedge \eta=\sum_{k,j=1}^{2g}\mu_j\nu_k \int_{M}\alpha_j\wedge\alpha_k=
\sum_{j=1}^{g}\mu_j\nu_{j+g} \int_{M}\alpha_j\wedge\alpha_{j+g}+
\sum_{j=g+1}^{2g}\mu_j\nu_{jg} \int_{M}\alpha_j\wedge\alpha_{j-g}=
\sum_{j=1}^{g}\mu_j\nu_{j+g} -
\sum_{j=g+1}^{2g}\mu_j\nu_{jg}=\sum_{i=1}^g \int_{a_i}\omega\int_{b_i}\eta-\int_{a_i}\eta\int_{b_i}\omega.$$
I hope this helps.
Try starting from the other direction and use
$(F^*\omega) \otimes (F^*\eta)(v_{\sigma(1)},\ldots, v_{\sigma(p+q)}) = (F^*\omega)(v_{\sigma(1)},\ldots,v_{\sigma(p)})\cdot (F^*\eta)(v_{\sigma(p+1)},\ldots, v_{\sigma(p+q)}) = \omega(F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p)}))\cdot\eta(F(v_{\sigma(p+1)}),\ldots,F(v_{\sigma(p+q)})) = \omega \otimes \eta (F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p+q)}))$
and then compare to what you already have.
Best Answer
The answer mentioned should have noted that we can say: \begin{equation} \begin{aligned} \varphi^*(\omega)(v_1, \dotsc, v_k)&\varphi^*(\eta)(v_{k+1}, \dotsc, v_{k+l}) \\&= \omega(dF^*(v_1), \dotsc, dF^*(v_k))\eta(dF^*(v_{k+1}), \dotsc, dF^*(v_{k+l})) \end{aligned} \end{equation} and we can extend this termwise result over the $\text{Alt}$ map.