My attempt
Using the van Kampen theorem, we get:
$$\pi_1 (U_1) \cong \pi_1 (S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z} = \langle \alpha, \beta \ | \ \varnothing \rangle$$
$$\pi_1 (U_2) = \langle a, b \ | \ abab^{-1} = e \rangle$$
$$\pi_1 (U_1 \cap U_2) \cong \pi_1 (S^1) = \mathbb{Z} = \langle \gamma \ | \ \varnothing \rangle$$
Now, if we want to find the relation $R_s$, we need to think about how to properly project an element of $\pi_1 (U_1 \cap U_2)$ onto $\pi_1 (U_1)$ and $\pi_1 (U_2)$. As we have for $\pi_1 (U_1 \cap U_2)$ a loop $\gamma$ here, we simply realize, that for $\pi_1 (U_1)$:
$$\gamma \mapsto \alpha \beta$$
and for $\pi_1 (U_2)$:
$$\gamma \mapsto abab^{-1}$$
This would mean, that our relation $R_s$ is simply $abab^{-1} = \alpha \beta$, so ultimately the fundamental group for a Klein bottle with two points removed is:
$$\pi_1 (B \backslash \{ x, y \}) = \langle \alpha, \beta, a, b \ | \ \varnothing, abab^{-1} = e, abab^{-1} = \alpha \beta \rangle$$
But this is a wrong solution apparently, as it should be equal to $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$ (as a Klein bottle minus two points is homeomorphic to a wedge sum of three $S^1$).
So what's wrong here?
Best Answer
Your calculation of $\pi_1(U_2)$ is wrong: the presentation should not have that relator, it should instead just be $\pi_1(U_2) = \langle a, b \mid \emptyset \rangle$, because $U_2$ deformation retracts onto the union of the $a$ and $b$ circles.
So when you remove $abab^{-1}=e$ from the relator list for $\pi_1(B \setminus \{x,y\})$, you should quickly get to a presentation for $\mathbb Z * \mathbb Z * \mathbb Z$.