This is a long comment, not an answer.

Let $E_k$ be the total space of the orientable bundle over $S^2$ with fiber $\mathbb R^2$ and Euler class $k$. $E_k = \mathbb R^2 \rtimes_k S^2$. Let $\pi : E_k \to S^2$ be the bundle projection.

$C_2 E_k = \{ (x,y) \in E_k^2 : x \neq y\}$ is the configuration space, with $p : C_2 E_k \to E_k$ the map $p(x,y)=x$.

Consider the composite of $\pi \circ p : C_2 E_k \to S^2$. It's a fibration and the fibers are homotopy-equivalent to $S^3 \vee S^2$, although that's not the most honest way of perceiving the fibers. The idea is to think of $\pi \circ p(x,y) = \pi(x)$ as a point in the $0$-section of $\pi$. If $\pi(y) \neq \pi(x)$ you homotope $y$ to $\pi(y)$, i.e. the 0-vector over $\pi(y)$. If $\pi(y) = \pi(x)$ you can't do this in $E_k$, so you can homotope $\pi(y)$ to be a unit vector in $\pi^{-1}(\pi(x))$. In other words, the fiber of $\pi \circ p$ over $\pi(x)$ looks like the sphere bundle of $\pi$ with all each circle fiber over points in a neighbourhood of $\pi(x)$ collapsed. So what's really going on is the fibers have as a deformation-retract a subspace that's $S^3$ union a $2$-cell, but the attachment map for the $2$-cell is along a great circle. The nice thing about this deformation-retract, is it's equivariant with respect to the monodromy. Precisely,

$$C_2 E_k \simeq (S^3 \cup e^2) \rtimes S^2$$

where the monodromy $SO_2 \to Aut(S^3 \cup e^2)$ is rotation about this great circle. Specifically, if you think of $S^3$ as the unit sphere in $\mathbb C^2$, and let the great circle be $S^1 \times \{0\} \subset S^3$, then its the action of $S^1$ on $S^3$ given by $(z, (z_1,z_2)) = (z_1,zz_2)$. The action is trivial on the $2$-cell attachment.

So as a space, it's $S^3 \rtimes S^2$ union a $D^2 \times S^2$ attached along the $S^1 \times S^2 \subset S^3 \rtimes S^2$ corresponding to where the monodromy is trivial.

I think the attaching map is null-homologous in $H_* (S^3 \rtimes_k S^2)$. So this means

$H_*(C_2 E_k)$ is free abelian, with ranks $1, 0, 2, 1, 1, 1, 0, 0, 0$ in dimensions 0 through 8 respectively. The $H_4(C_2 E_k)$ class is interesting, have you computed its self-intersection number?

arehomotopy invariants if the manifolds are highly connected and the number of points is small (open: just simply connected + many points), but S&L give a counterexample when the compact manifolds have fundamental group. – Ben Wieland Mar 27 '12 at 22:53twopoints is also still open. – Mark Grant Mar 28 '12 at 15:39