As I mentioned in my comment above, this is a special case of a more general result: if $F \to E \xrightarrow{p} B$ is a smooth fiber bundle, then the fibers of $\pi$ define a foliation of $E$ by leaves which are diffeomorphic to $F$. Let me first give the steps of how one proves this result.
- The fibers of $p$ are embedded submanifolds of $E$ by the regular value theorem.
- By local triviality, we can find a collection of charts $\{(U_i, \eta_i)\}$ which cover $B$, and diffeomorphisms $\psi_i : p^{-1}(U_i) \to F\times U_i$ which respect the two natural projections to $U_i$.
- As $F$ is a manifold, we can find a collection of charts $\{(V_j, \chi_j)\}$ which cover $F$.
- The desired charts on $E$ are $\{(\psi_i^{-1}(V_j\times U_i), \phi_{i,j})\}$ where $\phi_{i,j} = (\chi_j, \eta_i)\circ\psi_i$.
I will carry out these steps (and verify the claims made) in the explicit case you asked about, namely the Hopf fibration.
Consider the map $p : S^3 \to \mathbb{CP}^1$ given by $p(z, w) = [z, w]$; note that the preimages under this map are precisely the equivalence classes: $p^{-1}(p(z, w)) = \{e^{i\theta}(z, w) \mid \theta \in \mathbb{R}\} = [(z, w)]$. The map $p$ is a surjective submersion, so every point in $\mathbb{CP}^1$ is a regular value and hence every preimage is an embedded submanifold of $S^3$. This establishes property $(1)$.
Consider the open set $U_1 = \mathbb{CP}^1\setminus\{[1, 0]\}$ and the map $\psi_1 : p^{-1}(U_1) \to S^1\times U_1$ given by $(z, w) \mapsto (\frac{w}{|w|}, [z, w])$. This is a smooth map with smooth inverse $(e^{i\theta}, [a, b]) \mapsto \frac{|b|e^{i\theta}}{b\sqrt{|a|^2+|b|^2}}(a, b)$. That is, $\psi_1$ is a diffeomorphism. Moreover, if we denote the natural projection $S^1\times U_1 \to U_1$ by $\operatorname{pr}_2$, then we have
$$\operatorname{pr}_2(\Psi((z, w))) = \operatorname{pr}_2((\tfrac{w}{|w|}, [z, w])) = [z, w] = p((z, w)).$$
It follows that $[(z, w)]$ is mapped diffeomorphically to $S^1\times\{[z, w]\}$.
Note that the points $(i, [z, w]) \in S^1\times U_1$ are precisely those for which $\operatorname{arg}(w) = \frac{\pi}{2}$, so $\psi_1^{-1}((S^1\setminus\{i\})\times U_1) = p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \frac{\pi}{2}\}$. Each of the factors of $(S^1\setminus\{i\})\times U_1$ are diffeomorphic to Euclidean spaces. Explicitly, we have
diffeomorphisms $\eta_1 : U_1 \to \mathbb{R}^2$ given by $([z, w]) \mapsto (\operatorname{Re}(\frac{z}{w}), \operatorname{Im}(\frac{z}{w}))$ and $\chi_1 : S^1\setminus\{i\} \to \mathbb{R}$ given by $\lambda \mapsto \frac{\operatorname{Re}(\lambda)}{1-\operatorname{Im}(\lambda)}$; note that $\chi_1$ is precisely stereographic projection from the 'north pole'.
Putting what we have together, the map
$$\phi_{1,1} : p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \tfrac{\pi}{2}\} \to \mathbb{R}^3$$
given by $\phi_{1,1} := (\chi_1, \eta_1)\circ\psi_1$ is a diffeomorphism and hence viewed as a coordinate chart. Moreover, if $L_{\alpha} = [(z, w)]$, then $(p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \frac{\pi}{2}\})\cap L_{\alpha}$ is mapped diffeomorphically to $\mathbb{R}\times\{(\operatorname{Re}(\frac{z}{w}), \operatorname{Im}(\frac{z}{w})\}$. That is, if $\varphi_{1,1} = (x^1, x^2, x^3)$, the intersection of the leaf with the coordinate chart is given by the equations $x^2 = \operatorname{Re}(\frac{z}{w}), x^3 = \operatorname{Im}(\frac{z}{w})$.
Replacing $\chi_1 : S^1\setminus\{i\} \to \mathbb{R}$ with $\chi_2 : S^1\setminus\{-i\}$ given by $\lambda \mapsto \frac{\operatorname{Re}(\lambda)}{1+\operatorname{Im}(\lambda)}$ (stereographic projection from the 'south pole'), we obtain a similar chart
$$\phi_{1,2} = (\chi_2, \eta_1)\circ\psi_1 : p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \tfrac{3\pi}{2}\} \to \mathbb{R}^3.$$
Now let $U_2 = \mathbb{CP}^1\setminus\{[0, 1]\}$, then there is a diffeomorphism $\phi_2 : p^{-1}(U_2) \to S^1\times U_2$ respecting the two projections to $U_2$ given by $(z, w) \mapsto (\frac{z}{|z|}, [z, w])$. Replacing $\eta_1 : U_1 \to \mathbb{R}^2$ with $\eta_2 : U_2 \to \mathbb{R}^2$ given by $[z, w] \mapsto (\operatorname{Re}(\frac{w}{z}), \operatorname{Im}(\frac{w}{z}))$, we obtain two more charts
$$\phi_{2,1} = (\chi_1, \eta_2)\circ\psi_2 : p^{-1}(U_2)\setminus\{\operatorname{arg}(w) = \tfrac{\pi}{2}\} \to \mathbb{R}^3$$
and
$$\phi_{2,2} = (\chi_2, \eta_2)\circ\psi_2 : p^{-1}(U_2)\setminus\{\operatorname{arg}(w) = \tfrac{3\pi}{2}\} \to \mathbb{R}^3.$$
The domains of the four charts $\phi_{1,1}, \phi_{1, 2}, \phi_{2, 1}, \phi_{2,2}$ cover $S^3$ and satisfy the requirements of $(2)$.
Best Answer
Given an atlas $\Phi_i = (\Phi_i^1, \Phi_i^2)$ as you described, define the foliation locally by the preimage of the horizontal subspaces. The condition on the cocycles (transition functions) will imply that these local submanifolds glue together to make the leaves. In particular, for any point $p$ in the domain of a $\Phi_j$, then this chart send the foliation around $p$ to the foliation of horizontal subspaces. As seen in the following image
taken from page 22 of Camacho, Lins Neto - Geometric Theory of Foliations.
ADDED: As requested by the OP, we need to show how these leaves are in fact immersed submanifolds. It is not really trivial. I'll try to sketch the idea but for a precise proof I refer to Lemma 1.2.16 in Candel, Colon - Foliations I
We give another structure to $M$ using the given the atlas $(U_i,\Phi_i)$.
First we give $\mathbb{R}^k$ the standard topology and give $\mathbb{R}^{n-k}$ the discrete topology. Then consider $\mathbb{R}^n =\mathbb{R}^k\times \mathbb{R}^{n-k}$ with the product topology. This leads to a structure of $k$-dimensional manifold in $\mathbb{R}^n$.
By the condition on the transition functions (which says that $(U_i,\Phi_i)$ is a coherent foliated atlas) we can extend this sctucture to $M$. Then $M$ will be a $k$-dimensional manifold whose connected components are exactly the leaves of the foliation, here enters the problem with the 2nd countability.
Let $x\in M$ and consider the leaf $L_x$ defined by the equivalence relation $x\sim y$ if there is a finite chain of plaques joining $x and y$. The foliated atlas $(U_i,\Phi_i)$ is locally finite by the $n$-manifold structure of $M$. Then each plaque $\Phi_i^{-1}(\{y=c\})$ intersect only a finite number of neighboring plaques. Since each plaque is itself 2nd countable, $L_x$ will be 2nd countable as well.
Then both $L_x$ and $M$ are the $k$-dimensional manifolds with this new structure and the inclusion is naturally an smooth embedding. This amounts to the inclusion of $L_x$ in $M$ being an immersion if $M$ is considered with the original $n$-dimension structure.