Let us consider the map $z \to z^k$. Since its derivative vanishes at $z=0$, in view of the Inverse Function Theorem, $f$ is locally invertible at every point $z \neq 0$.
Hence, for each $w_0 \neq 0$ and each $z_0$ such that $z_0^k=w_0$, it is possible to find a holomorphic function $f_{z_0}^{-1}$ defined in a neighborhood of $w_0$ that is the local inverse of $z^k$. This is substantially a branch of the $k$-th root.
Now, let $z_0=1$ identifying a branch of $z=w^{\frac{1}{k}}$ in a neighborhood of $w_0=1$. My questions are:
What is the value of the limit $$\lim\limits_{t \to 1} f_{z_0}^{-1}(e^{2 \pi i t})?$$
It is $e^{\frac{2\pi i}{k}}$. How is this result related to the impossibility to extend in a continuous way the domain of the branch of the $k$-th rooth over the whole $\mathbb{C} \setminus \{0\}$?
Secondly, in which sense I must interpret the following sentence:
"If one follows a branch of the $k$-th root along a closed path that winds around the origin, one eventually comes back to a different branch of the $k$-th root."
Finally, if $g(w)$ is a holomorphic function from $\mathbb{C}$ to $\mathbb{C}$, how can I prove that for any $w_0$ such that $g(w_0)\neq 0$ there exists a neighborhood of $w_0$ and exactly $k$ distinct choices for a holomorphic map $\hat g$ such that $\hat g^k(w) =g(w)$?
As Lee Mosher pointed out in the comment section above, $$\lim_{t \to 1} f_{z_0}^{-1}(e^{2\pi it})=1$$ since $f_{z_0}^{-1}$ is continuous in a neighborhood $\mathcal{U}$ of $1$. But let me try to reformulate what I think your question is.
Let's try to define $f_{z_0}^{-1}$ on all of $\mathbb{C}$. We want to extend the definition to $$f_{z_0}^{-1}(re^{i\theta})=\sqrt[k]{r}e^{\frac{i\theta}k},$$ for $\theta \in [-2\pi,2\pi]$. Note that this definition agree with the definition of your inverse in a small neighborhood of $1.$
We want to show that $f_{z_0}^{-1}$ is not even continuous. Consider the path given by $e^{\pi i t}$ for $t \in [-1,1]$. Then by definition, $f_{z_0}^{-1}(e^{-\pi i}) = e^{\frac{-i\pi}{k}} \neq e^{\frac{i\pi}{k}} = f_{z_0}^{-1}(e^{\pi i}).$ This is a contradiction, since $ e^{-\pi i} =-1= e^{\pi i}.$ Therefore we showed that the function is not continuous at $-1$.
This follows from the fact that $f$ is a topological branched self-cover of the complex plane, branched at $0$. In particular, if you restrict the definition of $f$ to $\mathbb{C}-\{0\},$ then you get a honest cover. A loop around zero based at $z_0$ will lift to $k$ different path joining the $k$ pre-images of $z_0$. I think that this comment answer your second question on the common:
I claim that this is the only thing that can go wrong. If you can't take a loop around $0$, then $f_{z_0}^{-1}$ will extend to a holomorphic function (using the definition above). Therefore, the maximal domain of definition of $f_{z_0}^{-1}$ is $\mathbb{C} - \mathbb{R}_{\leq 0}.$ Note that there is nothing particular with the ray that we picked except the fact that it was based at $0$ and did not contain the point $z_0$ that we started with.
Now we can answer your last question:
First, note that for any point $z_0 \neq 0,$ you can define $$ f_{z_0,n}^{-1}(re^{i\theta})=\sqrt[k]{r}e^{\frac{i\theta+2\pi in}k},$$ give you a different holomorphic function in a neighborhood of $z_0$ for all $n \in \{ 0,1, \cdots, k-1\}.$ To conclude, consider $z_0=g(\omega_0)$ and let $\hat{g} = f_{g(\omega_0),n}^{-1} \circ g.$ By definition, $\hat{g}^k=f (\hat{g}) = g$ in a neighborhood of $\omega_0$.