So, I start off with this model:
It's clear to see that $U \cong S^1 \vee S^1$ and $U\cap V \cong S^1$. For $V$:
so $V\cong S^1$ also. So $$ \pi_1(T^2\setminus {\sim}) = \frac{\pi_1(S^1\vee S^1)*\pi_1(S^1)}{N} = \frac{\mathbb{Z}*\mathbb{Z}*\mathbb{Z}}{N} $$ now we need to deal with the inclusions to get $N$. For ${f : \pi_1(U\cap V) \to \pi_1(U)}$, it's we have $f(1) = aba^{-1}b^{-1}$ ($1$ is the generating loop in $U\cap V$). So basically ${\mathbb{Z}*\mathbb{Z}}$ get's abelianized to $\mathbb{Z}\times \mathbb{Z}$. But for $g : \pi_1(U\cap V)\to \pi_1(V)$, $g(1) = 1$, so one of the copies of $\mathbb{Z}$ cancels so we get $$ \pi_1(T^2\setminus {\sim}) = \mathbb{Z}\times \mathbb{Z} $$ However, the answer should be ${(\mathbb{Z}\times \mathbb{Z})*\mathbb{Z}}$ I think. What's wrong with my reasoning? I thought maybe $g(1)=1$ could be wrong, but I don't see how. A loop that goes around once in $U\cap V$ should be homotopy equivalent to the loop that goes around once in $V$?