All you need to know is that $S^{n-1}$ is a smooth submanifold of $\mathbb R^n$ and thereby receives a canonical smooth structure. It is well-known that this smooth structure is also generated by the smooth atlas given by your two stereographic projections $\chi^{n-1}_N$ and $\chi^{n-1}_S$ from north pole $N$ and south pole $S$. There are also other smooth atlases which are simpler, but have more charts. As an example take the $2n$ charts living on $U_i^\epsilon = \{ (x_1,\dots,x_n) \in S^{n-1} \mid (-1)^\epsilon x_i > 0 \}$, with $i = 1,\dots,n$ and $\epsilon =\pm 1$, and projecting $U_i^\epsilon$ onto the standard open $(n-1)$-disk by omitting the $i$-th coordinate.
Let us observe that the "stereographic atlas" $\mathcal S_{n-1}$ can be modified by choosing any linear automorphism $L : \mathbb R^{n-1} \to \mathbb R^{n-1}$ and defining $\mathcal S^L_{n-1} =\{L \circ \chi^{n-1}_N, \chi^{n-1}_S \}$. This atlas clearly gives the same smooth structure.
Now write $\mathbb R^4 = \mathbb C^2$ and $\mathbb R^3 = \mathbb C \times \mathbb R$. Then $S^3 \subset \mathbb C^2$ and $S^2 \subset \mathbb C \times \mathbb R$. Let us redefine the Hopf map $h : S^3 \to S^2$ by
$$h(z_0,z_1) = (2\bar z_0 z_1,\lvert z_0 \rvert^2 - \lvert z_1 \rvert^2 ). \tag{1}$$
Note that this differs from your definition by the coordinate permutation $\pi : S^2 \to S^2, \pi(x_1,x_2,x_3) = (x_3,x_2,x_1)$. The latter is a diffeomorphism (it is the restriction of the corresponding coordinate permutation on $\mathbb R^3$ to the submanifold $S^2$). Clearly our new $h$ is a smooth submersion iff the original $h$ is one. The purpose of this redefinition was to obtain the above elegant formula for $h(z_0,z_1)$.
Now define
$$\rho : \mathbb{CP}^1 \to S^2, \rho([z_0:z_1]) = \frac{1}{\lvert z_0 \rvert^2 + \lvert z_1 \rvert^2}(2\bar z_0 z_1,\lvert z_0 \rvert^2 - \lvert z_1 \rvert^2 ) . $$
This is a well-defined continuous map. Note that $\rho([1:0]) = N$. We shall show that it is a bijection, hence a homeomorphism because $\mathbb{CP}^1$ is compact and $S^2$ Hausdorff. Define
$$\sigma : S^2 \to \mathbb{CP}^1,\sigma(z,t) = \begin{cases} [\bar z:1-t] & (z,t) \ne N \\ [1:0] & (z,t) = N \end{cases}$$
Note that for $(z,t) \in S^2 \subset \mathbb R \times \mathbb R$ we have $t = 1$ iff $(z,t) = (0,1) = N$ = north pole of $S^2$. Also observe that each element of $\mathbb{CP}^1 $ has a representative of the form either $[z_0:1]$ or $[1:0]$. Using this it is easy to verify that $\sigma \circ \rho = id$ and $\rho \circ \sigma = id$.
On $S^2$ let us consider the smooth atlas $\mathcal S'_2 = \{L \circ \chi^2_N, \chi^2_S \}$, where $L$ is complex conjugation. It is easy to check that
$$L \circ \chi^2_N \circ \rho = \varphi_1 \quad, \quad \chi^2_S \circ \rho = \varphi_0 .$$
This shows that $\rho$ is a diffeomorphism. By construction of $\rho$ we have $\rho \circ H = h$. It therefore suffices to show that one of $H$ and $h$ is a smooth submersion.
For $h$ have a look at Proving that the Hopf map $S^3 \to S^2$ is a submersion.
For $H$ it is easier. Let $S^3_k = \{(z_0,z_1) \in S^3 \mid z_k \ne 0 \}$, $k= 0,1$. These are open subsets of $S^3$ which cover $S^3$. Consider the maps
$$H_k : S^3_k \stackrel{H}{\to} U_k \stackrel{\varphi_k}{\to} \mathbb C$$
which are given by $H_0(z_0,z_1) = \dfrac{z_1}{z_0}$ and $H_1(z_0,z_1) = \dfrac{z_0}{z_1}$. It suffices to show that these are submersions (smoothness is obvious). We shall do that for $H_0$, the other case is similar.
Given a point $(z_0,z_1) \in S^3_0$, define
$$\phi : \mathbb C \to \mathbb C^2, \phi(w) = \frac{1}{\lVert (z_0, z_0 w) \rVert}(z_0,z_0 w) .$$
This is a smooth map with image contained in $S^3_0$. Since $S^3_0$ is a submanifold of $\mathbb C^2$, the map
$$\psi : \mathbb C \stackrel{\phi}{\to} S^3_0$$
is smooth. We have $H_0 \circ \psi = id$ and therefore the differential of $H_0$ must be surjective in all points of $\psi(\mathbb C)$. But clearly $(z_0,z_1) = \psi(\dfrac{z_1}{z_0}) \in \psi(\mathbb C)$.
