Proving that the Hopf map $S^3 \to S^2$ is a submersion

There are three approaches:

1. Determine the image of $d_p\iota : T_pS^3 \to T_p\mathbb R^4$ and the kernel of $d_q \pi : T_q(\mathbb R^3 \setminus \{0\}) \to T_{\pi(q)} S^2$. [Note that $\pi$ is not defined in $0 \in \mathbb R^3$.] Then check $d_p(\pi F \iota)$. This should be easy, you can use the identifications $T_x\mathbb R^n = \mathbb R^n$ and $T_xS^{n-1} = \{ y \in \mathbb R^n \mid \langle x, y \rangle = 0 \}$ = orthogonal complement of $x$ in $\mathbb R^n$. Here $\langle -, - \rangle$ denotes the usual scalar product on $\mathbb R^n$.
   This is a good approach since you have already computed the Jacobian of $F$.

2. We have to check that $d_pH(T_pS^3) = T_{H(p)}S^2$, in other words that the image of $T_pS^3$ under $d_pH$ is two-dimensional. Now observe that $F \circ \iota = i \circ H$, where $i : S^2 \hookrightarrow \mathbb R^3$. Hence $$d_pF (d_p \iota(T_pS^3)) = d_{H(p)}i(d_pH(T_pS^3) . \tag{1}$$ Since $d_pF$ has rank $3$, its kernel $K$ is a one-dimensional subspace of $T_p\mathbb R^4$. Let $q : T_p\mathbb R^4 \to T_p\mathbb R^4/K$ be the quotient map. Clearly $d_pF$ induces an isomorphism $\phi : T_p\mathbb R^4/K \to T_{H(p)} \mathbb R^3$ and $$d_pF (d_p \iota(T_pS^3)) = \phi(q(d_p \iota(T_pS^3)). \tag{2}$$ It is well-known that $d_p\iota$ has rank $3$, i.e. $d_p \iota(T_pS^3)$ is a three dimensional subspace of $T_p\mathbb R^4$. Therefore $q(d_p \iota(T_pS^3))$ is at least two-dimensional and so is $d_pF (d_p \iota(T_pS^3))$. But now $(1)$ implies that $d_pH(T_pS^3)$ cannot have dimension less than $2$. This completes the proof.

3. Writing $\mathbb R^4 = \mathbb C^2$ and $\mathbb R^3 = \mathbb C \times \mathbb R$ we get $$F(w_0,w_1) = (2\bar w_0 w_1, \lvert w_0 \rvert^2 - \lvert w_1 \rvert^2) .$$ Note that for $\lambda \in S^1 \subset \mathbb C$ we have $F(\lambda((w_0,w_1)) = F(\lambda w_0, \lambda w_1) = F(w_0,w_1)$. Now define $$\phi : \mathbb C \times (-\infty,1) \to \mathbb C^2, \phi(w,t) = \frac{1}{\sqrt{2(1-t)}}(w,1-t) = \left(\frac{w}{\sqrt{2(1-t)}},\sqrt{\frac{1-t}{2}} \right).$$ This is clearly a smooth map. Its restriction to $S^2 \setminus \{(0,1)\}$ has the property $\lVert \phi(w,t) \rVert = \frac{\lvert w \rvert^2 + (1-t)^2}{2(1-t)} = \frac{\lvert w \rvert^2 + 1-2t +t^2}{2(1-t)} = \frac{2-2t}{2(1-t)} = 1$, i.e. gives a smooth map $\phi' : S^2 \setminus \{(0,1)\} \to S^3$. It is easy to verify that $$H(\phi'(w,t)) = (w,t) .$$ Clearly for each $\lambda \in S^1$ the map $\lambda \phi'$ is smooth and we have $H(\lambda\phi'(w,t)) = (w,t)$. Thus $\lambda\phi'$ is a smooth section of $H$ on $S^2 \setminus \{(0,1)\}$ which implies that $T_pH$ is surjective for all $p$ in the image of some $\lambda\phi'$. But each $(w_0,w_1) \in S^3$ with $w_1 \ne 0$ is in the image of $\frac{w_1}{\lvert w_1 \rvert} \phi'$:
   It suffices to show that each $(z_0,t_1) \in S^3$, where $t_1$ is a real number in $(0,1]$, is in the image of $\phi'$. In fact, then we know that $(\lvert w_1 \rvert \frac{w_0}{w_1}, \lvert w_1 \rvert) \in S^3$ is in the image of $\phi'$ and hence $(w_0,w_1)$ is in the image of $\frac{w_1}{\lvert w_1 \rvert} \phi'$. So let $(z_0,t_1)$ be as above. Then $(2t_1z_0, 1-2t_1^2) \in S^2 \setminus \{(0,1)\}$ and $\phi(2t_1z_0, 1-2t_1^2) = (z_0,t_1)$.
   For the points $(w_0,0) \in S^3$ we argue as follows. Let $U : \mathbb C^2 \to \mathbb C^2, U(w_0,w_1) = (w_1,w_0)$, be the coordinate switch map which is a diffeomorphism. It restricts to a diffeomorphism $U' : S^3 \to S^3$. We have $$HU'(w_0,w_1) = (2\bar w_1 w_0, \lvert w_0 \rvert^2 - \lvert w_1 \rvert^2) = G'H(w_0,w_1),$$ where $G : \mathbb C \times \mathbb R\to \mathbb C \times \mathbb R, G(w,t) = (\bar w, -t)$ is a diffeomorphism which restricts to a diffeomorphism $G' : S^2 \to S^2$. Thus $G'H U' = HU'U' = H$ and we conclude that the rank of $T_pH$ agrees with the rank of $T_{U(p)}H$. But for the points $(w_0,0)$ we have $U(w_0,0) = (0,w_0)$ and this case was treated above.