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Let us consider the square $I \times I$ and define on its boundary the following equivalence relation: $(0,t) \sim (1,t)$ for any $t \in I$ and $(\frac{1}{2},0) \sim (\frac{1}{2},1)$. Which is the fundamental group of $X=\partial (I \times I)/\sim$?

I argue as follows: $X$ is a cylinder with two points identified (the centers of the circular bases). I can consider two arcs $A$ and $B$ joining the previous points but $B$ all contained in the cylinder and $A$ intersecting the cyilinder only at the endpoints. In this way, it should be $\pi_1(X)=\pi_1((S^1 \times \mathbb{R}) \vee S^1)=\mathbb{Z} * \mathbb{Z}$. Is my argument right or not?

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Yes. Alternatively, consider two lines in the fundamental square. The first connects $(1/2,0)$ to $(1/2,1)$ and the other connects $(0,1/2)$ to $(1,1/2)$. The image of the union of these lines under the identifications given is clearly $S^1\vee S^1$, and we can also see that $X$ deformation retracts onto this image, so $\pi_1(X)=\pi_1(S^1\vee S^1)\cong \mathbb{Z}*\mathbb{Z}$.

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    The argument given by the OP is not correct. They could be some non-trivial relations in the fundamental group. – 2017-01-03
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    @N.H. I agree. I looked too hastily to the OP's final answer. Thanks for pointing it out. – 2017-01-03
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Your argument just show that there is a surjection $F_2 \to \pi_1(X)$ but this is not a complete argument.

But you can easily retract (even deform retract) $X$ on $S^1 \vee S^1$ which has $F_2 = \mathbb Z * \mathbb Z$ as fundamental group, so the final answer was correct.