How do I informally prove this using Van Kampen theorem? 
Let $X$ be the space obtained from two tori $S^1\times S^1$ by identifying a circle $S^1\times\{x_0\}$ in one torus with the corresponding circle $S^1\times\{x_0\}$ in the other. Calculate $\pi_1(X)$.

Well, my professor only explains what van-Kampen theorem is about (not in detail though) and handed a class an exercise and he's gonna put this in exam tomorrow.
So.. I'm bad at informal things.. How do I show this? Could someone please help..?
I think the problem is asking to calculate this 
Where should I cut?
 A: Let $\alpha_1$ and $\beta_1$ be loops on $(T^2)_1$ which generate $\pi_1((T^2)_1)$ and $\alpha_2$ and $\beta_2$ be the corresponding loops in $(T^2)_2$. The space $X$ is what we get when we glue the tori together alone $\alpha_1$ and $\alpha_2$.
Let $U_1\subset X$ be a small open neighbourhood of $(T^2)_1$ in $X$ and similarly let $U_2\subset X$ be a small open neighbourhood of $(T^2)_2$ in $X$. The intersection $U_1\cap U_2$ is a small open neighbourhood of $\alpha_1=\alpha_2$ and so in particular has fundamental group which is generated by these elements. We can conclude from Van-Kampen's theorem that $$\pi_1(X)=((\mathbb{Z}^2\ast\mathbb{Z}^2)_{\langle \alpha_1,\beta_1,\alpha_2,\beta_2 \rangle})/\langle\alpha_1=\alpha_2\rangle$$

Honestly, the much quicker way to calculate this fundamental group, without worrying about Van-Kampen's theorem, is to see that $X$ is homeomorphic to the space $(S^1\vee S^1)\times S^1$, the product of a circle with a wedge of two circles (figure eight space). We can then use the fact that $\pi_1(A\vee B)\cong \pi_1(A)\ast \pi_1(B)$ and $\pi_1(A\times B)\cong \pi_1(A)\times\pi_1(B)$ to get $\pi_1(X)\cong (\mathbb{Z}\ast\mathbb{Z})\times\mathbb{Z}$ - here the three generaters from left to right are $\beta_1,\beta_2,\alpha_1(=\alpha_2)$.
A: Take $A$ to be one torus with title neighborhoods around the other torus and $B$ to be another torus with little neighborhoods around the another torus. $A \cup B$ then is the whole space $X$ and $A \cap B$ deformation retracts onto a circle.
So $\pi_1(X) \cong \pi_1(A) \star \pi_1(B)/\langle i_A^{-1}, i_B \rangle$ which is isomorphic to $\Bbb Z^2 \star \Bbb Z^2$ with the identification $(1, 0) \sim (1, 0)$. The fundamental group then is $\langle a, b, c, d |[a, b] = [c, d] = ac^{-1} = 1 \rangle$.
A: Let $U_1, U_2$ be the tori, and then $U_1\cap U_2$ is a circle. Recall that $\pi_1(U_i)=\mathbb Z^2$ and $\pi_1(S^1)=\mathbb Z$.
By VKT, the fundamental group is $\mathbb Z^2*\mathbb Z^2/\sim$, where $\sim$ identifies $(1, 0)$ in the first $\mathbb Z^2$ with $(1, 0)$ in the other. So we should get $\langle a, b, c, d\mid[a, b]=[c, d]=ac^{-1}=1\rangle=\langle a, b, d \mid [a, b]=[a, d]=1\rangle$.
