Almost always, these are two different actions.
The monodromy action is by path lifting, lifting a given path at each of the points in the fiber and mapping each such point to the endpoint of the corresponding lift. This makes sense for any covering space $q:(Y,y_0) \to (X,x_0)$, where $Y$ is possibly disconnected.
To get a well-defined action by the deck group $G=G(\widetilde{X})$, which can be identified with $\pi_1(X,x_0)$ in the universal cover $p:(\widetilde{X},\widetilde{x}_0) \to (X,X_0)$, on the fiber of an intermediate connected normal cover $q:(Y,y_0) \to (X,x_0)$, certain elements need to identified: $G(Y) \cong G/N$, where $N= q_*\pi_1(Y,y_0)$. So, there is an action of $G$ on $F = q^{-1}(x_0)$ that factors through the action of $G/N$ on $F$.
Here's an example where these two actions are different:
Let $X = S^1 \vee S^1$ with oriented labeled edges $x$ and $y$, identified with generators of $\pi_1(X)$. And let $Y = \widetilde{X}$ be its universal cover, viewed as a graph (4-valent tree) with labeled oriented edges and its vertex set identified with $\pi_1(X)$.
Consider the pair of adjacent vertices $(1,y)$. The monodromy action by the element $x$ sends this pair to $(x,yx)$, which are not adjacent in $Y$.
The deck group action by the element $x$ sends the pair to $(x,xy)$.
In general, the deck group action extends to a homeomorphism of $Y$. But the monodromy action need not extend continuously.
The above example is related to other actions: if $Y$ is a normal covering graph with fundamental group $N \trianglelefteq G$, where $G$ is a free group, then $Y$ is the right Cayley graph of the presentation $G/N$. The action of $F/N$ by deck transformations is the left action (by graph automorphisms). The monodromy action is the action on the vertex set by right multiplication (which does not extend to an action by graph automorphisms).
Exercise #27 in Chapter 1 of Hatcher's Algebraic Topology book is worth a look.
Suppose that $X$ is a complete (connected) Riemannian manifold, $\Gamma$ is a group of isometries of $X$ acting on $X$ properly discontinuously.
Definition. Let $\gamma\in\Gamma$ be an element of infinite order. Then a complete geodesic $A_\gamma\subset X$ is called an {\em axis} of $\gamma$ if it is invariant under $\gamma$.
(In this case, $\gamma$ acts on $A_\gamma$ as a translation. Note that an axis need not be unique.)
Given an infinite order element $\alpha\in\Gamma$, let $X_\alpha$ denote the quotient Riemannian manifold $X/\langle \alpha\rangle$.
I will be always parameterizing geodesics by the arc-length.
Lemma 1. Suppose that $\alpha, \beta\in\Gamma$ are elements of infinite order so that the subgroups $\langle \alpha\rangle$, $\langle \beta\rangle$ have trivial intersection. Then the projection $A_\beta\to X_\alpha$ is a proper map.
Suppose not. Then there exists a sequence of points $z_i\in A_\beta$ diverging to infinity such that the projections of these points to $X_\alpha$ are within distance $R$ from some point $\bar{x}\in X_\alpha$ (projection of $x\in X$).
Since $\langle \beta\rangle$ acts cocompactly on the axis
$A_\beta$, there is a constant $C$ and a diverging sequence of integers $n_i$ such that the (minimal) distance from $\beta^{n_i}(x)$ to the $\langle\alpha\rangle$-orbit of $x$ is $\le C$. In other words, there is a diverging sequence of integers $(m_i)$ such that
$$
d(\beta^{n_i}(x), \alpha^{-m_i}(x))\le C.
$$
Without loss of generality, we may assume that both sequences consist of distinct integers. We have
$$
d(\alpha^{n_i} \beta^{n_i}(x), x)\le C
$$
By the proper discontinuity of the action of $\Gamma$ on $X$, the set of products
$\alpha^{n_i} \beta^{n_i}$ is finite. Hence, there exist distinct $i, j$ such that
$$
\alpha^{n_i} \beta^{n_i}= \alpha^{n_j} \beta^{n_j},
$$
$$
\alpha^{n_i-n_j}= \beta^{n_j-n_i},
$$
which is a contradiction. qed
Suppose now that $\Gamma$ satisfies the following maximality condition: If $\alpha, \beta$ are primitive elements of $\Gamma$ then the subgroups generated by $\alpha, \beta$ are either equal or have trivial intersection. For instance, fundamental groups of surfaces satisfy this property. (Let me know if you need help proving this property. For noncompact surfaces this follows from the fact that their fundamental groups are free.)
Corollary 1. Suppose that $\alpha\in \Gamma$ is a primitive element. Then for each $\gamma\in\Gamma$ $\alpha, gamma$ either commute or anticommute:
$$
[\alpha,\gamma]=1, \hbox{or} \gamma\alpha\gamma^{-1}=\alpha^{-1},
$$
or the geodesic $\gamma(A_\alpha)$ projects properly to $X_\alpha$.
Proof. Consider
$$
\beta= \gamma\alpha\gamma^{-1}.
$$
Then $A_\beta=\gamma(A_\alpha)$. By Lemma 1, if the projection of $A_\beta$ to $X_\alpha$ is not proper, then the subgroups generated by $\alpha, \beta$ have nontrivial intersection. By the Maximality Property above, these subgroups have to be equal (note that the primitivity of $\alpha$ implies that of $\beta$ since conjugation is a group automorphism). Thus,
$$
\langle \alpha\rangle= \langle\gamma\alpha\gamma^{-1}\rangle.
$$
Since the infinite cyclic group $\langle \alpha\rangle$ has only two generators, $\alpha^{\pm 1}$, we get
$$
[\alpha,\gamma]=1, \hbox{or} \gamma\alpha\gamma^{-1}=\alpha^{-1},
$$
qed
Suppose now that $X$ is 2-dimensional and $\Gamma$ acts freely. If $\Gamma$ preserves orientation, i.e. $M=X/\Gamma$ is orientable, the anti-commutation is impossible. Furthermore, unless the group $\Gamma$ is isomorphic to ${\mathbb Z}^2$ and the quotient $X/\Gamma$ is diffeomorphic to the torus, the commutation is only possible if $\gamma$ belongs to the subgroup generated by $\alpha$.
Thus, we obtain:
Corollary 2. Suppose that $X$ is 2-dimensional, $M=X/\Gamma$ is orientable and is not diffeomorphic to the torus. Then either $\gamma(A_\alpha)=A_\alpha$ or $\gamma(A_\alpha)$ projects properly to $X_\alpha$.
Note that the torus is a genuine exception: No matter what $\gamma\in \Gamma$ you take, $\gamma(A_\alpha)$ does not project properly to $X_\alpha$ since
$\gamma(A_\alpha)$ is invariant under $\alpha$ and, hence, projects to a compact in $M_\alpha$.
Best Answer
Take $g \in \pi_{1}(X,x_0)$ and $a,b \in \widetilde{X}$. We would like to show that $\tilde{d}(a,b) = \tilde{d}(g(a),g(b))$.
Now, if $\gamma$ join $a$ and $b$ then $g(\gamma)$ joins $g(a)$ and $g(b)$. Conversely, if $\gamma$ joins $g(a)$ and $g(b)$, then $g^{-1}(\gamma)$ joins $a$ and $b$. Let us write $\Gamma(a,b)$ for the set of paths from $a$ to $b$, and $\Gamma(g(a),g(b))$ for the set of paths from $g(a)$ to $g(b)$. We see that $\gamma \mapsto g(\gamma)$ is a bijection between $\Gamma(a,b)$ and $\Gamma(g(a),g(b))$.
Next, by the definition of Deck transformations, $p(\gamma) = p(g(\gamma))$ for any path $\gamma$. Thus $p(\Gamma(a,b)) = p(\Gamma(g(a),g(b)))$, i.e. these two sets of paths have the same projection in $X$.
Finally, $\tilde{d}(a,b) = \inf_{\gamma \in p(\Gamma(a,b))} \ell(\gamma) = \inf_{\gamma \in p(\Gamma(g(a),g(b)))} \ell(\gamma) = \tilde{d}(g(a),g(b))$, completing the proof.