Fundamental Group of a Quotient under Group Action

You need to assume more about the action than just freeness; otherwise this statement is not true. For example, $\mathbb R$ acts freely on itself by translation, and $\mathbb R$ is simply connected, but $\mathbb R/\mathbb R$ is a single point, and its fundamental group is not isomorphic to $\mathbb R$.

In order to get the conclusion you want, you also have to assume that the action satisfies the following condition:

Every $x \in X$ has a neighborhood $U$ such that for all $g\in G$, $gU \cap U = \emptyset$ unless $g$ is the identity.

(Note that this implies, in particular, that $G$ acts freely.) Some authors call an action satisfying this condition a "properly discontinuous action"; but I don't like that term because different authors use it with inequivalent definitions, and also because it leads to oxymoronic expressions like "a continuous properly discontinuous action." Allen Hatcher in his book Algebraic Topology introduced the term covering space action for a continuous group action satisfying the condition above, and I've adopted that term in my books.

The basic fact is that if the action of $G$ on $X$ is a covering space action, then the quotient map $X\to X/G$ is a covering map. If in addition $X$ is simply connected, then $\pi_1(X/G)$ is isomorphic to $G$.

For more details, see this MSE answer; and for even more, see Chapters 11 and 12 in my book Introduction to Topological Manifolds.