I don't understand the kernel of $\Phi$ in Van Kampen's theorem. If we have $A_1$, ..., $A_n$, $X=A_1\cup\cdots\cup A_n$ and $\Phi:\pi_1(A_1,x_0)\ast\cdots\ast \pi_1(A_n,x_0)\rightarrow \pi_1(X$,x_0) being the extended homomorphism then the Van Kampen theorem states that:
$\pi_1(X,x_0)=\frac{\pi_1(A_1,x_0)\ast\cdots\ast \pi_1(A_n,x_0)}{ker(\Phi)}$
with
$ker(\Phi)=\langle\langle [f]_{A_i}[f]_{A_j}^{-1} \Big| [f]\in\pi_1(A_i\cap A_j\rangle\rangle$
I'm trying to understand, in words, what the group that is the $ker(\Phi)$ is.
For the free product of the fundamental groups, I get that. Its just saying "take loops in the individual $A_i$ and then do them one at a time to your heart's content and that would be a loop in $X$."
But, for the kernel, is it just saying "go ahead and take thoe loops, but remember, if you take a loop in one $A_i$ and then a loop in an $A_j$, check to see if they are actually, secretly, opposites of each other in $A_i\cap A_j$ and, if they are, you can just replace that part with the identity loops (i.e. remove it)."?
If that's right, how does this work in practice? For example, with the double Torus here, I can see how the $\pi_1(U)=\pi_1(V)=\mathbb{Z}\ast\mathbb{Z}$ and so the overall free product becomes $\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$. But, $U\cap V$ is the shell of a cylinder without a top or bottom and so, to me, it looks like $ker(\Phi)$ is just any loop around the center and then going backwards. But I'm not able to put it together to "see" what the overall fundamental group is.
 A: This kernel is chosen to allow the identification of specific elements in the overall free product. Namely, if you start from the more general setting where $A,B,C$ are groups with two morphisms $\varphi:C\to A$ and $\psi:C\to B$, then you can think of $\varphi(C)$ as a subgroup of $A$ (thus of $A\ast B$) and of $\psi(C)$ as a subgroup of $B$ (thus of $A\ast B$). If you want to formally identify these two subgroups in $A\ast B$, then you have to quotient it by the normal subgroup $N$ generated by the elements of the form $\varphi(c)\psi(c)^{-1}$, $c\in C$, since you will then have in the quotient $(A\ast B)/N$ that
$$[\varphi(c)]=[\varphi(c)\varphi^{-1}(c)\psi(c)]=[\psi(c)]$$
for all $c\in C$. This group is usually denoted
$$(A\ast B)/N=A\underset{C}{\ast} B.$$
What Van Kampen's theorem tells you is that (under the assumption that $X=U\cup V$ where $U,V$ and $U\cap V$ are path-connected open sets of $X$, $x_0\in U\cap V$):
$$\pi_1(X,x_0)\simeq\pi_1(U,x_0)\underset{\pi_1(U\cap V,x_0)}{\ast}\pi_1(V,x_0),$$
where the corresponding morphisms $\varphi$ and $\psi$ are induced by the inclusions $i:U\cap V \to U$ and $j:U\cap V\to V$.
In your example of the double torus, $\pi_1(U,x_0)\simeq \mathbb{Z}\ast\mathbb{Z}$ is free generated by two loops $a$ and $b$, $\pi_1(V,x_0)\simeq \mathbb{Z}\ast\mathbb{Z}$ is free generated by two loops $c$ and $d$, and $\pi_1(U\cap V,x_0)\simeq \mathbb{Z}$ is free generated by a loop $e$ which checks that $\varphi(e)=aba^{-1}b^{-1}=[a,b]$ and $\psi(e)=cdc^{-1}d^{-1}=[c,d]$, so $\pi_1(X,x_0)$ will have the presentation which is given in the answer you linked, namely
$$\pi_1(X,x_0)\simeq\langle a,b,c,d\,|\,[a,b][c,d]^{-1}\rangle.$$
