Sorry, this isn't a full answer, rather some algebraic hints at how you might relate the cross product to your intuition about the cup product (wrt intersections) and to linear algebra. It should really be a comment but it's too long.
First, a quick review! The cross product, $\times$, is actually part of the cup product:
$$x \smile y := \Delta^*(x \times y)$$
The diagonal embedding $X \xrightarrow{\Delta} X \times X$, is simply a canonical way to embed a space X into an ambient space endowed with the product topology, $\Delta X := \{(x,x) \in X \times X\}$. It is useful when want to look in the neighborhood of a space $X$ (e.g., at germs of functions on $X$), but $X$ sits in no ambient space. The word, "diagonal embedding," comes from the example of embedding of $R^1 \hookrightarrow R^2$ taking $x \mapsto (x,x)$, that is, taking the line $R^1$ and embedding it into $R^2$ as the line $y=x$.
This is the reason we have a cup product in cohomology and not in homology. The map induced by $\Delta$ on homology, $H_*(X) \xrightarrow{\Delta_*} H_*(X \times X)$, being the "wrong direction" for a product (from 1 to 2), whereas the induced map on cohomology, $H^*(X \times X) \xrightarrow{\Delta^*} H^*(X)$, is the right direction (from 2 to 1), and lends itself to the following precomposition.
The cup product (where $p+q=k$),
$$H^k(X) \longleftarrow H^p(X) \times H^q(X)$$
is actually
$$H^k(X) \leftarrow H^k(X \times X) \leftarrow H^p(X) \times H^q(X)$$
What is this mysterious map $H^p(X) \times H^q(X) \to H^{p+q}(X \times X)$? It's called the "cross product", $\times$, and, as you know, is defined more generally on $H^p(X) \times H^q(Y) \to H^{p+q}(X \times Y)$.
The cross product relates the cohomology groups of two different spaces to the cohomology groups of their product space.
I also think in pictures, so I understand the desire for a cartoon picture, but the algebra is actually quite enlightening here.
The tensor product of two graded abelian groups is $$(A^* \otimes B^*)^n := \bigoplus_{i+j=n} A^i \otimes B^{\text{ }j}$$ which, if you prefer thinking in terms of smash rather than tensor, will remind you of how we get from an $i$-cell in a pointed CW-complex X and a $j$-cell in a pointed CW-complex in Y, to an $i+j$-cell in $X \wedge Y$.
$$(X \wedge Y)_{n} := \times_{i+j=n}(X_i \wedge Y_j)$$
Returning our gaze to the cross product: $$H^p(C^*) \times H^q(D^*) \xrightarrow{\times} H^{p+q}(C^* \otimes D^*)$$
where $\alpha(w_i)$ is 0 when $w_i$ and $\alpha$ are of different degrees, similarly with $\beta(z_i)$.
This might strike you as a bit strange: how does this definition of the cross product match up with yours? Let's revisit Hatcher's definition in terms of the map induced by the projections of the Cartesian product.
$$X \leftarrow X \times Y \rightarrow Y$$
$$H^*(X) \rightarrow H^*(X \times Y) \leftarrow H^*(Y)$$
Taking the pullback:
where the group labeled by the notational atrocity $H^*(X) \times_{H^*(X \times Y)} H^*(Y)$ is defined as the friendly pairs of objects $\{(a,b) \text{ such that } p_1^*(a) = p_2^*(b)\}$ for all $a \in H^*(X), b \in H^*(Y)\}$.
Hatcher's definition of $a \times b$ is then:
$$H^*(X \times Y) \times H^*(X \times Y) \to H^*(X \times Y)$$
$$\text{( }p_1^*(a), p_2^*(b) \text{ )} \mapsto p_1^*(a) \smile p_2^*(b)$$
So, recall that I defined the cup product as $p_1^*(a) \smile p_2^*(b) := \Delta^*(p_1^*(a) \times p_2^*(b))$. Well, wait a minute, how do we define the diagonal embedding on a $X \times Y$, simple: $\Delta(X \times Y) := {(x,y) \text{ such that } x = y} = X \cap Y$, the diagonal embedding is the embedding $X \cap Y \hookrightarrow X \times Y$ (which, since $X \cap X = X$, reduces to the usual diagonal embedding).
Assuming that the definitions of cup product are equivalent, $a \times b = p_1^*(a) \smile p_2^*(b) = \Delta^*(p_1^*(a) \times p_2^*(b))$. From the point of view of the cup product, now have a hint as to what is going on: we're shoving $a$ and $b$ into the same range s.t. we can use our intuition about the cup product (which is only defined for X = Y) to recon with them.
Some other comments:
Bott says something like: a cocycle is a creature which lurks over spaces, pounces on cycles, eats them, and spits out numbers. In other words, it's a lot like a linear functional. In fact, I encourage you to recall that maps $f$ between modules satisfy the same rules that linear maps do, e.g., $f(ax +y) = af(x) + f(y)$. Indeed, this notation, i.e., $(C^k, d^k):= (Hom_R(C_k, R), d_k^*)$, a bit abhorrent at first glance, is actually telling us something. It's just borrowing from familiar linear algebra, for example, for a linear map between R-vector spaces $f: V \to W$, when we look at the dual spaces $V^*:=Hom_{R-Vect}(V, R)$ and $W^*:=Hom_{R-Vect}(V, R)$, we take the transpose $f^*: W^* \to V^*$.
The cross product, I assume, gets its name from being a generalization of the cross product of vectors in $R^3$ we know and love. For any two 1-forms, $\alpha$ and $\beta$, their wedge product is equivalent to the Hodge dual of their cross product:
$$\alpha \wedge \beta = *(\alpha \times \beta)$$
This is only true for 1-forms, but in general, the wedge product of p-form and a q-form is a p+q-form. I suspect that we might picture the cross product on cohomology groups via looking at simplicial groups rather than chain complexes, but I don't understand them well enough to describe how we might take their cohomology without geometrically realizing them or passing back to chain complexes anyway.
Best Answer
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.