Let $X$ be a path connected, locally path connected space and let $p:\tilde{X} \to X$ be a covering map. Let $x_0\in X$. Then we have a natural right action of $\pi_1(X, x_0)$ on the fibre $p^{-1}(x_0)$ given by $x'_0\cdot [\gamma] =\tilde{\gamma} (1)$. Where $\tilde {\gamma} $ is the unique lift of $\gamma$ beginning at $x'_0$.
However, is there an action of this group on all of $\tilde {X}$?
On
NB: My "answer" here mistakenly assumes that $\tilde{X}$ denotes the universal cover, which is simply connected. In other words, it gives the details for the claim made in @user316092's answer, the one that starts with "But if $\tilde X$ is the universal cover..." Because it's a useful explanation of that part, I decided to leave it here rather than removing it.
Sort of. But you have to pick a particular point in the fiber over $x_0$, say $x_0'$. Now for any other point $a$ of the covering space, there's a unique (up to homotopy rel endpoints) path $\alpha$ from $x_0$ to $a$, which leads to a path $\beta = p \circ \alpha$ in $X$, from $x_0$ to $p(a)$.
For $\gamma$ as above, let $\zeta = \gamma \cdot \beta$, a path in $X$ from $x_0$ to $p(a)$. Now $\zeta$ has a unique lift to $\hat{\zeta}$, a path in $\tilde{X}$ starting at $x_0'$. Then define
$$ F_\gamma (a) = \hat{\zeta}(1) $$
Note that for $a$ in the fiber, this degenerates to the definition you described.
As you noted in your question, $\pi_1(X,x_0)$ acts on $p^{-1}(x_0)$ by permuting the points (i.e. the lifts of $x_0$) around. But if you're looking for a group which acts on the entire space $\tilde X$ (which I believe is your question?) then you'd want the group of deck transformations $G(\tilde X)$. This group is the collection of all covering space isomorphisms $f:(\tilde X, \tilde x_0)\to (\tilde X,\tilde x_0)$, i.e. homeomorphisms of $\tilde X$ which satisfy $p=p\circ f$. (So intuitively, a homeomorphism $f$ is an element of $G(\tilde X)$ if the projection or "shadow" of $\tilde X$ onto $X$ looks just like the projection/"shadow" of the 'homeomorphed' space $f(\tilde X)$ onto $X$).
In general, $\pi_1(X)$ and $G(\tilde X)$ are not isomorphic. But if $\tilde X$ is the universal cover then they are. (See, for instance, Proposition 1.39 of Hatcher to see a more general relationship between $G$ and $\pi_1$.) So in that case, the action of $\pi_1$ on the elements of $p^{-1}(x_0)$ does indeed coincide with an action on all of $\tilde X$.