These are indeed not the same in most cases (including when $X=S^1\vee S^1$ or $X=S^1\times S^1$) for the rather trivial reason that the first action is defined using $\bar{\gamma}$ and the second is defined using $\gamma$ (so at least when acting on the basepoint $\tilde{x}_0$, the action of $[\gamma]$ by the first definition corresponds to the action of $[\gamma]^{-1}$ by the second). I'm guessing that what Hatcher had in mind, though, is the version of the first action which is defined using a lift of $\gamma$, not a lift of $\bar{\gamma}$. Of course, that makes the first action a right action rather than a left action, so the actions cannot possibly be the same when $\pi_1(X,x_0)$ is nonabelian (given that the actions are faithful). But for an abelian group, right and left actions are the same, and so this still leaves a nontrivial question of whether they coincide when $\pi_1(X,x_0)$ is abelian.
Strictly speaking, it doesn't make sense to say the monodromy action is equal to $\rho$ (even up to conjugation) as they are two actions on different sets. The correct notion of equivalence is isomorphism of group actions, defined as follows. If $G$ acts on both $X$ and $Y$, the two actions are isomorphic if there is a bijection $f: X \to Y$ such that $g\cdot f(x) = f(g\cdot x)$. You should convince yourself that this definition describes the notion of two actions being "the same".
If both sets are numbered, we sometimes conflate the actions with the isomorphic actions on $\{1,\ldots, d\}$, whereupon action isomorphism turns into conjugation. This can sometimes confuse matters, so let's be explicit: we are looking for a bijection between the fiber $\pi^{-1}(x)$ and $\{1,\ldots, d\}$ which commutes with the two actions of $\pi_1(V,q)$ (respectively, the monodromy action and the action $\rho$, given by $g\cdot x = \rho(g)(x)$).
The high-level explanation is that both the monodromy action and $\rho$ are isomorphic to the action of $\pi_1(V,q)$ on the cosets of $H$ (given by $b\cdot aH = (ba)H$).
The monodromy action is isomorphic to the action on cosets.
I will compose paths right to left, so that $ah \in aH$ is first $h$, then $a$. Then the lift of $ah$ based at $\tilde{q}$ (which I denote $\widetilde{ah}_{\tilde{q}})$ is the concatenation of $\tilde{h}_{\tilde{q}}$ (a loop in $\tilde{V}$) and $\tilde{a}_{\tilde{q}}$. In particular, the lift of every $ah \in aH$ based at $\tilde{q}$ has the same endpoint, namely $\tilde{a}_{\tilde{q}}(1) \in \pi^{-1}(q)$. This gives a function $f:G/H \to \pi^{-1}(q)$ by $f(aH) = \tilde{a}_{\tilde{q}}(1)$. Any $y \in \pi^{-1}(q)$ can be connected to $\tilde{q}$ by a path $\tilde{\gamma}$, and $f(\pi(\tilde{\gamma}) H) = y$, so $f$ is moreover bijective.
$b \in \pi_1(V,q)$ acts on $x \in \pi^{-1}(q)$ by sending it to $\tilde{b}_x(1)$. When $x = f(aH) = \tilde{a}_{\tilde{q}}(1)$, we get $$b\cdot f(aH) =b\cdot \tilde{a}_{\tilde{q}}(1) = \tilde{b}_{\tilde{a}_{\tilde{q}}(1)}(1),$$ which is a complicated way of saying "first lift $a$ based at $q$, then lift $b$ based at the endpoint of $\tilde{a}_{\tilde{q}}$, and take the endpoint of that". But by uniqueness of lifts, this is the same as just taking the endpoint of $\widetilde{ba}_{\tilde{q}}$. Then $$b\cdot f(aH) = \tilde{b}_{\tilde{a}_{\tilde{q}}(1)}(1) = f((ba)H) = f(b\cdot aH).$$ Then the two actions are isomorphic, as claimed.
The action on cosets is isomorphic to $\rho$.
The final piece of the puzzle is pure group theory—if $\rho: G \to S_n$ is a transitive* action (via $g\cdot x = \rho(g)(x)$), then it is isomorphic to the action on the cosets of the stabilizer $H = \text{Stab}(1)$. The isomorphism is given by $j(aH) = a \cdot 1$, and I'll leave it to you to verify it is indeed an isomorphism of actions (it's essentially a much easier version of the above).
Then $j\circ f^{-1}: \pi^{-1}(q) \to \{1,\ldots,d\}$ is an isomorphism between the monodromy action of $\tilde{V}$ and $\rho$.
*If $\rho$ is not transitive, then the statement is false, as the monodromy action of a connected covering space is always transitive.
Best Answer
$\bar{M}$ is the set of curves $\alpha:[0,1]\to M, \alpha(0)=x_0$ quotiented by $\alpha\sim \alpha_2$ if $\alpha(1)=\alpha_2(1)$ and $\alpha\cup \alpha_2^-$ is equal in $\pi_1(M,x_0)$ to the trivial curve $Tr(t)=x_0$.
The action of $\pi_1(M,x_0)$ on $\bar{M}$ is $\gamma\cdot\alpha = \gamma\cup \alpha$.
As $M$ is a connected manifold then $\bar{M}$ is a covering space. $\bar{M}$ is simply connected because a closed-loop $f:Tr\to Tr$ in $\bar{M}$ is represented by a continuous function $f:[0,1]\times [0,1]\to M$ such that $f(t,0)=f(0,t)=f(1,t)=x_0$, each $f(t,.)$ is an element of $\bar{M}$, and the continuous function $g:[0,1]\times [0,1]\times [0,1]\to M$, $g(r,t,s)=f((1-r)t,s)$ represents an homotopy from $f$ to the trivial loop at $Tr$.