Double torus minus an open disk Why is the fundamental group of the double torus minus an open disk given by $\mathbb{Z} \star \mathbb{Z} \star \mathbb{Z} \star \mathbb{Z}$? 
More specifically, why do we need to delete the open disk?
 A: The genus-2 surface (I assume this is what you mean by "double torus"), when punctured, deformation-retracts onto its 1-skeleton, which is a "bouquet" (wedge sum) of 4 circles. The fundamental group of the bouquet of circles is the free group $\Bbb{Z} \ast \Bbb{Z} \ast \Bbb{Z} \ast \Bbb{Z}$, and since a deformation retraction is a homotopy equivalence, the fundamental group of the punctured surface is the same. 
If the surface is not punctured, this deformation retract is not possible. In this case, you get extra relations, with some of the generators commuting with each other. See Wikipedia's article on van Kampen's Theorem for a more detailed explanation/proof of this example.
A: If you consider the fundamental polygon of the double torus, which is given by an octagon with the appropriate identifications, then a deformation retraction is given by simply the edges of this octagon (no interior). My allowing the appropriate identifications, you arrive at the wedge sum of four circles; that is, four circles joined at a single point. By applying the van Kampen theorem (I am using Hatcher's Algebraic Topology for reference): - 
Theorem 1.20. If $X$ is the union of path-connected open sets $A_{\alpha}$ each containing the base point $x_0 \in X$ and if each intersection $A_{\alpha} \cap A_{\beta}$ is path-connected, then the homomorphism $\Phi : \ast_{\alpha} \pi_1(A_{\alpha}) \to \pi_1(X)$ is surjective. If in addition each intersection $A_{\alpha} \cap A_{\beta} \cap A_{\gamma}$ is path-connected, then the kernel of $\Phi$ is the normal subgroup generated by all elements of the form $i_{\alpha \beta}(\omega) i_{\beta \alpha}(\omega)^{-1}$ for $\omega \in \pi_1(A_{\alpha} \cap A_{\beta})$,and hence $\Phi$ induces an isomorphism $\pi_1(X) \cong \ast_{\alpha} \pi_1(A_{\alpha})/N$. 
you arrive at $\pi_1(X) \cong \mathbb{Z} \ast \mathbb{Z} \ast \mathbb{Z} \ast \mathbb{Z}$. $\Box$
Hatcher's text is given here: https://www.math.cornell.edu/~hatcher/AT/AT.pdf
I'm not sure what you mean by your second question. 
