To see what is going on, I think it is helpful to carefully work out an explicit example. In Section 2.1 of Calegari's paper, he explains how to deal with 3-manifolds that fiber over the circle. Here is a brief description of what he does.

I will start by setting up notation (and I won't follow Calegari here since I prefer to write it a little differently). Let $\Sigma_g$ be a compact oriented genus $g$ surface and let $\psi\colon \Sigma_g \rightarrow \Sigma_g$ be an orientation-preserving diffeomorphism. Let $M_{\psi}$ be the mapping torus of $\psi$, so $M_{\psi} = \Sigma_g \times [0,1] / \sim$ where $(p,1) \sim (\psi(x),0)$ for all $x \in \Sigma_g$. The fibers of the map $M_{\psi} \rightarrow S^1$ are all genus-$g$ surfaces, and form a taut foliation.

This points to one place where I think you slightly misunderstand what Calegari does. It's easy to find a transversal to the above taut foliation that hits each fiber exactly once. But you can't ensure that this is the entire branch locus and that the branching is simple since there might not be a branched cover $\Sigma_g \rightarrow S^2$ with a single branch point of order $2$. For what he is doing, you'll have to take a more complicated transversal with several components.

Let $P \subset S^2$ be a set of $4g$ distinct points. We can then easily construct a map $\pi\colon \Sigma_g \rightarrow S^2$ that is a degree $(g+1)$ branched cover, where the branch locus is exactly $P$ and each point of $p$ is a simple branch point (i.e., has degree $2$).

Now comes the key step in Calegari's argument. He argues that we can find an orientation-preserving diffeomorphism $\phi\colon S^2 \rightarrow S^2$ taking $P$ to $P$ such that $\psi\colon \Sigma_g \rightarrow \Sigma_g$ is a lift of $\phi$, i.e., such that
$$\pi(\psi(x)) = \phi(\pi(x)) \quad \quad \text{for all $x \in \Sigma_g$}.$$
Letting $M_{\phi}$ be the mapping torus of $\phi$, we thus get a map
$F\colon M_{\psi} \rightarrow M_{\phi}$ that is a fiberwise branched cover.

Calegari stops the argument here, and omits the final step of why this gives a map $f\colon M_{\psi} \rightarrow S^2$ that is a branched cover on each fiber. Here is what I think he has in mind. The mapping class group of $S^2$ is trivial, so the map $\phi\colon S^2 \rightarrow S^2$ is isotopic to the identity. Of course, this isotopy moves the points in $P$. This implies that there is a fiber-preserving diffeomorphism $\lambda\colon M_{\phi} \rightarrow M_{\text{id}} = S^2 \times S^1$. The map $f\colon M_{\psi} \rightarrow S^2$ is then
$$M_{\psi} \stackrel{F}{\longrightarrow} M_{\phi} \stackrel{\lambda}{\longrightarrow} S^2 \times S^1 \stackrel{\text{proj}}{\longrightarrow} S^2.$$
One feature of this is that it is not the case that the branch points in the fibers all map to the same points in $S^2$; indeed, as you go around the circle of fibers the images of branch points in $S^2$ trace out a braid. I think it would hard to avoid this.

Since the composition $\lambda \circ F$ is fiber-preserving, your $1$-dimensional foliation I think will be exactly the standard foliation of the mapping torus $M_{\psi}$ whose leaves are unions of the images of sets $x \times [0,1]$ in $M_{\psi}$. If $x$ is a periodic point of $\psi$, then this leaf will be a closed circle, but for non-periodic points it will be $\mathbb{R}$.