?
If Rudin regarded $\Phi$ as a function of $2$-forms and suppose that $\omega=f(\mathbf{x})dx_{i_1}\land dx_{i_2}$ is $2$-form on $\mathbb{R}^m$ then $$I_{\Phi}(\omega)=\int \limits_{\Phi}\omega=\int \limits_{I^2} f(\Phi(\mathbf{u}))\begin{vmatrix} D_1\phi_{i_1} & D_2\phi_{i_1} \\ D_1\phi_{i_2} & D_2\phi_{i_2} \end{vmatrix}(u_1,u_2)d\mathbf{u}=\int \limits_{\sigma_1(Q^2)}+\int \limits_{\sigma_2(Q^2)}$$ Let's take a look at first integral: $$\int \limits_{\sigma_1(Q^2)}f(\Phi(\mathbf{u}))\begin{vmatrix} D_1\phi_{i_1} & D_2\phi_{i_1} \\ D_1\phi_{i_2} & D_2\phi_{i_2} \end{vmatrix}(u_1,u_2)d\mathbf{u}$$ Making transform $\mathbf{u}\mapsto \sigma_1(\mathbf{u})$ and since $\sigma_1$ is injective and $C'$-mapping (and it's Jacobian is $1$) then by theorem 10.9 we have $$\int \limits_{Q^2}f(\Phi(\sigma_1(\mathbf{u})))\begin{vmatrix} D_1\phi_{i_1}\circ \sigma_1 & D_2\phi_{i_1}\circ \sigma_1 \\ D_1\phi_{i_2}\circ \sigma_1 & D_2\phi_{i_2}\circ \sigma_1 \end{vmatrix}(u_1,u_2)d\mathbf{u}$$ But for right application of theorem 10.9 support of integrand must be compact and lies in $\sigma_1(Q^2)$. Right? But we have no information about it.
The second question why $\Phi\circ \sigma_1+\Phi\circ \sigma_2$ is $2$-chain?
I don't have Rudin at hand, but in the hope a picture is worth a thousand words: You presumably have a standard $2$-simplex $Q^{2}$, and are partitioning the unit square $I^{2}$ as shown:
If $\omega$ is a $2$-form defined on the image $\Phi(I^{2})$, the integral breaks into two summands, one for each simplex, each an integral over the standard simplex.
A $2$-chain is a formal (finite) integer linear combination of images of the standard $2$-simplex, i.e., an element of the free Abelian group on the space of $2$-simplices. The diagram should make obvious the sense in which $$ \Phi(I^{2}) = \Phi \circ \sigma_{1}(Q^{2}) + \Phi \circ \sigma_{2}(Q^{2}) $$ as $2$-chains.