Why is the correlation coefficient always smaller than $1$ and bigger than $-1$? I kind of see that the covariance is similar to the product of the standard deviations but I can't figure out how. My problem is that I don't understand how does this formula tells us how strong the connection is between 2 variables.

In addition to the other excellent answers, I want to offer an explanation using the Cauchy-Schwarz inequality which states:
$$
|\langle \mathbf{u},\mathbf{v}\rangle|^2 \leq \langle \mathbf{u},\mathbf{u}\rangle \cdot \langle \mathbf{v},\mathbf{v}\rangle
$$
where $\langle\cdot,\cdot\rangle$ is the inner product. After defining an inner product on the set of random variables using the expected value of their product,
$$
\langle X,Y\rangle := \operatorname{E}(XY)
$$
the Cauchy-Schwarz inequality becomes
$$
|\operatorname{E}(XY)|^{2} \leq \operatorname{E}(X^{2}) \operatorname{E}(Y^{2}).
$$
No we apply this to $(X-\mu_{x})$ and $(Y-\mu_{y})$:
$$
|\underbrace{\operatorname{E}((X-\mu_{x})(Y-\mu_{y}))}_{= \operatorname{Cov}(XY)}|^{2} \leq \underbrace{\operatorname{E}((X-\mu_{x})^{2})}_{= \operatorname{Var}(X)} \underbrace{\operatorname{E}((Y-\mu_{y})^{2})}_{= \operatorname{Var}(Y)}.
$$
Consequently,
$$
|\operatorname{Cov}(XY)|^{2}\leq \operatorname{Var}(X)\operatorname{Var}(Y)
$$
Taking square roots, we get
$$
|\operatorname{Cov}(XY)|\leq \sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)} = \sigma_{x}\sigma_{y} \implies -\sigma_{x}\sigma_{y}\leq \operatorname{Cov}(XY)\leq +\sigma_{x}\sigma_{y}.
$$
By dividing by $\sigma_{x}\sigma_{y}$ we get the correlation coefficient $\rho$:
$$
-1\leq \underbrace{\frac{\operatorname{Cov}(XY)}{\sigma_{x}\sigma_{y}}}_{=\rho}\leq 1.
$$
