$W=B(x_{1},r)\cap B(x_{2},r)$.
The boundary of the intersection is given by the union of
$\delta_{1}W=\delta B(x_{1},r)\cap B(x_2,r)$ and $\delta_{2}W=B(x_1,r)\cap \delta B(x_{2},r)$.
Let $W_{1}=\{sx_{1}+(1-s)z,0\le s\le 1, z\in\delta_{1}W\}$ and $W_{2}=\{sx_{2}+(1-s)z,0\le s\le1, z\in\delta_{2}W\}$.
Would someone be able to draw/visualise $W_{1}$ and $W_{2}$ for me?
Here is a picture:
The set $W$ is the intersection of the two circular regions. The red curve is $\delta_1 W$, and the green area is $W_1$. In fact, $W_1$ is the convex hull of $\{x_1\} \cup \delta_1W$. In words, it's the sector of the circle $B(x_1,r)$ subtended by the arc $\delta_1 W$.
Similarly, $W_2$ is the sector of the circle $B(x_2,r)$ subtended by the arc $\delta_2 W$.