What is the induced orientation on a product of vector spaces in singular cohomology?

This is an incredibly annoying subtlety. To unwind, we have to review how cup products work. Let $X$ be a topological space and $A,B\subseteq X$ be subspaces. In its most primitive form, the cup product is a map $C^i(X)\times C^j(X)\rightarrow C^{i+j}(X)$. This restricts to a cup product on relative cochains $C^i(X,A)\times C^j(X,B)\rightarrow C^{i+j}(X,A+B)$. On the other hand, there is an inclusion map $C^{i+j}(X,A\cup B)\rightarrow C^{i+j}(X,A+B)$. Now, all of these maps are compatible with the differential in an appropriate sense, hence induce maps in cohomology $H^i(X,A)\times H^j(X,B)\rightarrow H^{i+j}(X,A+B)$ and $H^{i+j}(X,A\cup B)\rightarrow H^{i+j}(X,A+B)$. The important feature now is that if $A$ and $B$ are open in $A\cup B$, then this latter map is an isomorphism. Composing with its inverse yields the true relative cup product on cohomology, $H^i(X,A)\times H^j(X,B)\rightarrow H^{i+j}(X,A\cup B)$. The important subtlety is that this inverse $H^{i+j}(X,A+B)\rightarrow H^{i+j}(X,A\cup B)$ is not given by the identity on representatives.

If we now apply this discussion to the scenario at hand, you will see that the cocycle $p_1^{\ast}\eta\cup p_2^{\ast}\eta$ represents an element of $H^2(\mathbb{R}^2,\mathbb{R}\times\mathbb{R}\setminus\{0\}+\mathbb{R}\setminus\{0\}\times\mathbb{R})$, but it does not represent an element of $H^2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$ directly. I leave it as an exercise to find an explicit $2$-chain (a $2$-simplex will do, actually) in $\mathbb{R}^2\setminus\{0\}$ on which $p_1^{\ast}\eta\cup p_2^{\ast}\eta$ does not vanish. Nonetheless, the cohomology class $[p_1^{\ast}\eta\cup p_2^{\ast}\eta]\in H^2(\mathbb{R}^2,\mathbb{R}\times\mathbb{R}\setminus\{0\}+\mathbb{R}\setminus\{0\}\times\mathbb{R})$ defines a cohomology class in $[\eta]\times[\eta]\in H^2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$, which we can pair with the generator $[\sigma]\in H_2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$. The downside is that we don't have an explicit representative of $[\eta]\times[\eta]$, which begs the question how we can actually concretely calculate this evaluation. Let me address this.

I will first go the long route and sketch a geometric way of understanding the inverse $H^{\bullet}(X,A+B)\rightarrow H^{\bullet}(X,A\cup B)$, which will explain how to do this calculation. Then I will give a formal argument. The chain map $C^{\bullet}(X,A\cup B)\rightarrow C^{\bullet}(X,A+B)$ is dual to the quotient map $C_{\bullet}(X,A+B)\rightarrow C_{\bullet}(X,A\cup B)$, which fits into a short exact sequence of chain complexes $$0\rightarrow C_{\bullet}(A\cup B,A+B)\rightarrow C_{\bullet}(X,A+B)\rightarrow C_{\bullet}(X,A\cup B)\rightarrow0.$$ The complex $C_{\bullet}(A\cup B,A+B)$ is acyclic (this is the excision theorem), whence the other map induces isomorphisms in homology. Geometrically, this acyclicity comes from the fact that a cycle in $A\cup B$ possesses a subdivision, which is a cycle in $A+B$. Similarly, an inverse $H_{\bullet}(X,A\cup B)\rightarrow H_{\bullet}(X,A+B)$ can be obtained by taking a relative cycle in $(X,A\cup B)$, i.e. a chain with boundary in $A\cup B$, and subdivide it (which subdivides the boundary as well, since subdivision is a chain map) until the boundary is a chain with boundary in $A+B$, whence the subdivided cycle represents a homology class in $(X,A+B)$. Lastly, the desired inverse $H^{\bullet}(X,A+B)\rightarrow H^{\bullet}(X,A\cup B)$ is obtained by the dual of this on representatives. This is only a sketch as I'm being intentionally vague about how exactly we should subdivide each chain, etc..

Formally, the point is that the maps $H^{\bullet}(X,A\cup B)\rightarrow H^{\bullet}(X,A+B)$ and $H_{\bullet}(X,A+B)\rightarrow H_{\bullet}(X,A\cup B)$ are adjoints with respect to the pairings $H^{\bullet}(X,A+B)\times H_{\bullet}(X,A+B)\rightarrow\mathbb{Z}$ and $H^{\bullet}(X,A\cup B)\times H_{\bullet}(X,A+B)\rightarrow\mathbb{Z}$ (which is a purely algebraic fact about chain maps and their duals).

Thus, to evaluate $[\eta]\times[\eta]\in H^2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$ on $[\sigma]\in H_2(\mathbb{R}^2,\mathbb{R}^2\setminus\{0\})$, we choose a homology class $\alpha\in H_2(\mathbb{R}^2,\mathbb{R}\times\mathbb{R}\setminus\{0\}+\mathbb{R}\setminus\{0\}\times\mathbb{R})$ that maps to $[\sigma]$ under the canonical map (this can be achieved by appropriately subdividing a representative) and evaluate on it the cocycle $p_1^{\ast}\eta\cup p_2^{\ast}\eta$.

I leave it to you to carry this out explicitly. An appropriate subdivision of $\sigma$ will involve at least four simplices, so might be a bit fidgety. Alternatively, you could consider an appropriate subdivision of the square $[-1,1]^2$ into two simplices, which will yield an easier calculation. Of course, you can also feel free to do both calculations and confirm they yield the same result.