This is well known Seifert-van Kampen theorem for which I am asking any help in order to make a clear picture of it.
Now, in most of my "first" exercises I'm working in such a way that I am usually able to find "only" two such open, path-connected subsets of $X$, usually denoted by $U$ and $V$.
I am given a space $X$ to be $X:=\mathbb D^2/\sim$ where $\sim$ stands as a relation such that $x\sim ix$ for each $x\in\mathbb S^1$.
My idea is to take for one set a small nbh. of $0$ and for another the complement of zero.
Okay, for the first case I know that the image of the generator of $\pi_1$ of the intersection is trivial since the other space is contractible to a point,
but what confuses me is how can I "read" what is the image of this generator in the other
and, at the end, how to conclude what my final fundamental group is?
Can I suggest you view your topological space in a slightly different way?
Take a unit circle, $$S^1 = \{ w \in \mathbb C : |w| = 1 \}.$$ Also take a closed unit disk, $$D^2 = \{ z \in \mathbb C : |z| \leq 1 \}.$$ Glue the boundary circle $\partial D^2$ of the disk $D^2$ to the circle $S^1$ by identifying each $z \in \partial D^2$ with the point $w = z^4 \in S^1$.
Thus $$ X = (S^1 \sqcup D^2)/\sim$$ where $$w \in S^1 \ \sim \ z \in D^2 \ \ \iff \ \ |z| = 1 { \rm \ and \ } w = z^4.$$
Can you convince yourself that this is the same space as what you described?
First, I recommend that we try to guess the fundamental group of this space using only our visual intuition! The idea is very simple: If we have a loop that wraps once, twice or three times around the $S^1$, then the loop is non-trivial. But if we have a loop that wraps four times around the $S^1$, then the loop can "slip off" into the interior of the disk $D^2$, and contract. So the fundamental group should be $\mathbb Z_4$.
Now, let's prove this intuition is correct using Van Kampen. Take $U$ to be an open neighbourhood of the $S^1$, $$ U = \{ z \in \mathbb C : 1-\epsilon_1 < |z| \leq 1 \}.$$ Take $V$ to be the interior of the $D^2$, $$ V = \{ z \in \mathbb C : |z| < 1-\epsilon_2 \},$$ where $\epsilon_2 < \epsilon_1$.
So what does the overlap region $U \cap V$ look like? In $z$ coordinates, we can write this overlap region as $$ U \cap V = \{ z \in \mathbb C : 1-\epsilon_1 < |z| < 1-\epsilon_2 \}.$$ This is an annulus. It shouldn't be hard to see that the fundamental group of $U \cap V$ is generated by the loop $\gamma_{U \cap V}$, defined by $$ z = r e^{2\pi i t}, \ \ \ \ \ t \in [0,1],$$ where $r$ is any radius between $1-\epsilon_1$ and $ 1-\epsilon_2$.
Now let us think about $\gamma_{U \cap V}$ as a loop in $U$ (rather than only as a loop in $U \cap V$). By expanding it to radius $1$, you can see that $\gamma_{U \cap V}$ is homotopic to the loop $$ z = e^{2\pi i t}, \ \ \ \ \ t \in [0,1]. $$ Since we identify $w \in S^1$ with $z \in \partial D^2$ when $w = z^4$, this loop can also be written as $$ w = e^{8\pi i t}, \ \ \ \ \ t \in [0,1]. $$
Since $U$ deformation retracts onto the circle $S^1$, the fundamental group of $U$ is generated by the loop $\gamma_U$, defined by $$ w = e^{2\pi i t}, \ \ \ \ \ t \in [0,1].$$
Thus, $\gamma_{U \cap V} \sim 4 \gamma_{U}$ in $\pi_1(U)$.
Meanwhile $V$ is a open disk, which has trivial fundamental group. So $\gamma_{U \cap V} \sim 0$ in $\pi_1(V)$.
Finally, Van Kampen tells you that $\pi_1(X)$ is generated by $\gamma_U$, except that the element $4\gamma_U$ should be identified with the element $0$. This group is precisely $\mathbb Z_4$.