= Solution
Given <group homomorphisms> $i_A:C\to A$ and $i_B:C\to B$, their <amalgamated free product> is a group $A*_CB$ with maps $j_A:A\to A*_CB$ and $j_B:B\to A*_CB$ satisfying $j_Ai_A=j_Bi_B$, such that every pair $f_A:A\to H$, $f_B:B\to H$ satisfying $f_Ai_A=f_Bi_B$ factors through a unique homomorphism $A*_CB\to H$.
The <Seifert-van Kampen theorem> says that if $X=U_1\cup U_2$, where $U_1,U_2$, and $U_1\cap U_2$ are open and path-connected and $x_0\in U_1\cap U_2$, then the inclusion maps induce
$$
\pi_1(X,x_0)\cong
\pi_1(U_1,x_0)*_{\pi_1(U_1\cap U_2,x_0)}
\pi_1(U_2,x_0).
$$
Here is a direct proof of the requested generation statement. For a <based loop> $\gamma$ in $X$, the <Lebesgue number lemma> supplies a subdivision
$$
0=t_0<t_1<\cdots<t_m=1
$$
such that every arc $\gamma|_{[t_{j-1},t_j]}$ lies in one $U_{\varepsilon_j}$. Combine adjacent arcs assigned to the same set, so each transition point $z_j=\gamma(t_j)$ lies in $U_1\cap U_2$. Since the intersection is <path-connected space>[path-connected], choose a path $q_j$ in it from $x_0$ to $z_j$, taking $q_0=q_m$ constant. Then
$$
q_{j-1}*\gamma|_{[t_{j-1},t_j]}*\overline{q_j}
$$
is a loop in $U_{\varepsilon_j}$. In the product of their classes, each $\overline{q_j}*q_j$ cancels by <path reversal>, leaving $[\gamma]$. Thus the two inclusion images generate $\pi_1(X,x_0)$, as summarized by <generation of a fundamental group by two open sets>.
Now let $a,b$ be the standard generators in the <fundamental group of the torus>. A <Möbius band> retracts to a core circle, while its boundary traverses that core twice. If $x$ and $y$ denote the core classes of the two attached bands, the <Seifert-van Kampen theorem> gives
$$
\pi_1(Y)
\cong
\langle a,b,x,y\mid [a,b]=1,\ x^2=ab,\ y^2=a^2b^3\rangle.
$$
The two attaching classes $A=ab$ and $B=a^2b^3$ commute. Their exponent matrix has determinant
$$
\det\begin{pmatrix}1&1\\2&3\end{pmatrix}=1,
$$
so they form a basis of $\langle a,b\rangle\cong\mathbb Z^2$, with
$$
a=A^3B^{-1},\qquad b=A^{-2}B.
$$
Substituting $A=x^2$ and $B=y^2$ eliminates $a,b$ and leaves the two-generator one-relator presentation
$$
\boxed{\pi_1(Y)\cong\langle x,y\mid[x^2,y^2]=1\rangle.}
$$
This is the calculation in <two Möbius bands attached to a torus>.
Finally send $x$ to $(12)$ and $y$ to $(23)$ in the <symmetric group> $S_3$. Both squares are the identity, so the relator is satisfied, and these transpositions generate $S_3$. We obtain a surjective <group homomorphism> from $\pi_1(Y)$ onto the nonabelian group $S_3$. Therefore $\pi_1(Y)$ is nonabelian.
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