On Wikipedia it says
A map $f: X\to Y$ is a quotient map if it is surjective, and a subset $U$ of $Y$ is open if and only if $f^{-1}(U)$ is open.
So by definition, if $f^{-1}(U) \subset X$ is open, then $U \subset Y$ is open. But I also read that a quotient map is not necessarily an open map. I saw some counter-example maps, but they were too difficult for me to understand intuitively. But if a map is not open, then there exists an open set $f^{-1}(U) \subset X$ such that $U \subset Y$ is not open. But by definition of quotient map, this is impossible, so I don't understand how this can be.
$f : X \to Y$ be your quotient map, $U \subset X$ be an open set and $f(U)$ be the image of $U$ in $Y$.
By definition, $U$ is open iff $f^{-1}(f(U))$ is open. Note here that $f^{-1}(f(U))$ needn't be the same as $U$, which is probably the source of your confusion.
For example, consider the map $\Bbb R \to S^1$ given by $x \mapsto e^{2\pi i x}$, where $S^1$ is realized as a subspace of $\Bbb C$. $[a, b]$ be an arbitrary small ($|a - b| < 1$, say) interval in $\Bbb R$. Image of $[a, b]$ via $f$ is an arc in $S^1$, while preimage of that arc is a union of translates of $[a, b]$, that is, $$f^{-1}(f([a, b])) = \bigcup_{n \in \Bbb Z} [a + n, b + n] \neq [a, b]$$
That said, $f$ is open iff for every $U \subset X$, $f^{-1}(f(U))$ is open.
For an example of a quotient map which is not open, consider the quotient map $f : \Bbb R \to \Bbb R/\Bbb Q$, $\Bbb R$ having the standard topology. Consider the open set $(1, -1) \in \Bbb R$. $f^{-1}(f((1, -1))) = (1, -1) \cup \Bbb Q$. This is not open in $\Bbb R$, hence $f$ cannot possibly be an open map.