= Solution
The <Seifert-van Kampen theorem> says that if $X=U\cup V$, where $U,V,U\cap V$ are path connected <open sets> containing $x_0$, then
$$
\pi_1(X,x_0)\cong
\bigl(\pi_1(U,x_0)*\pi_1(V,x_0)\bigr)\big/
\langle\!\langle i_{U*}(\gamma)i_{V*}(\gamma)^{-1}:
\gamma\in\pi_1(U\cap V,x_0)\rangle\!\rangle.
$$
The cell attachment is the quotient
$$
X\cup_fD^n=(X\sqcup D^n)/(u\sim f(u)\text{ for }u\in\partial D^n),
$$
with the image of $x_0=f(*)$ as base point. For $n=2$, take one <open set> that deformation retracts onto $X$ together with a boundary collar, and another that consists of the interior of the disc with a collar and is contractible. Their intersection deformation retracts onto $S^1$. Its generator maps to $[f]$ in the first set and to the identity in the second. Van Kampen therefore proves the <fundamental group after attaching a 2-cell> formula
$$
\boxed{\pi_1(X\cup_fD^2,x_0)
\cong\pi_1(X,x_0)/\langle\!\langle[f]\rangle\!\rangle}.
$$
Applying van Kampen to the two circles gives
$$
\pi_1(S^1\vee S^1,*)\cong F(a,b).
$$
Attach three discs along loops representing $a^2$, $b^3$, and $(ab)^2$. The resulting <presentation complex for the symmetric group on three letters> has <group>
$$
\langle a,b\mid a^2=b^3=(ab)^2=1\rangle.
$$
Sending $a$ to $(12)$ and $b$ to $(123)$ gives a surjection onto $S_3$. The relations imply $aba=b^{-1}$, so every word reduces to one of
$$
1,b,b^2,a,ab,ab^2.
$$
The presented <group> has at most six elements and surjects onto the six-element <group> $S_3$, so this map is an isomorphism.
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