Let be this picture: 
We try to figure out the fundamental group using the Van Kampen theorem.
Let p be a point on the center of the figure S, $U_1 = S-p$, $U_2 = S$ - "edge of figure", $U_0 = U_1 \cap U_2$. Then:
Amalgamated product $\pi_1(S) = \pi_1(U_1)*_{\pi_1(U_0)} \pi_1(U_2) =\mathbb{Z}*_{\mathbb{Z}} 0$. Then we have the morphism $h_1: \mathbb{Z} \longrightarrow \mathbb{Z}$ and $h_2: \mathbb{Z}\longrightarrow 0$. Clearly $h_2 \equiv 0$, but there is infinite $h_1$. I think $\mathbb{Z}*_{\mathbb{Z}} 0 = 0$. But could it be...? $$\mathbb{Z}*_{\mathbb{Z}} 0 = \mathbb{Z}/3\mathbb{Z}$$ Because it could be three revolutions, as I see in the figure "ababab".
Why do you think that taking an amalgamated product with $0$ gives $0$? What that operation does is identify the image of multiplication by $\mathbb Z\longrightarrow \mathbb Z$ in $\mathbb Z$ with $0$, so it is indeed the quotient you describe.
Concretely, if you have groups $G_1$ and $G_2$ and maps $f_2 : G_2\longleftarrow H\longrightarrow G_1 : f_1$, then the amalgamated product $G_1\ast_H G_2$ is the quotient of the free product $G_1\ast G_2$ by the relation that identifies the elements in $G_2$ coming from $f_2$ with the elements in $G_1$ coming from $f_1$.
In your case, we have that $\mathbb Z\ast 0 = \mathbb Z$ and that you are identifying $3\mathbb Z$ with $0$.