First we need some identities, using the conventions for $su(2)$ from the question:
\begin{align} [[A,B],C] &= 2A\ \mathrm{Tr}(BC) - 2B\ \mathrm{Tr}(AC), \\ [[A,B],[C,D]] &= 2A\ \mathrm{Tr}(C[D,B]) - 2B\ \mathrm{Tr}([A,C]D). \end{align}
If we take a derivative of $\mathrm{Tr}(\hat{\Phi}^2) = -2$, we find that $\mathrm{Tr}(\hat{\Phi}\ \partial_\mu\hat{\Phi}) = 0$. Using the first identity we find:
\begin{align} [[\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}], \hat{\Phi}] = 0 \implies [\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}] &\propto \hat{\Phi} \\ \implies [\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}] &= - \frac{1}{2} \mathrm{Tr}([\partial_\mu\hat{\Phi}, \partial_\nu\hat{\Phi}] \hat{\Phi}) \ \hat{\Phi}, \end{align}
where the constant of proportionality is determined by tracing against $\hat{\Phi}$.
Now, take the commutator of $D_\mu \hat{\Phi} = 0$ with $\hat{\Phi}$:
\begin{align} [\partial_\mu \hat{\Phi}, \hat{\Phi}] &= - [[A_\mu, \hat{\Phi}], \hat{\Phi}] \\ &= -2 A_\mu\ \mathrm{Tr}(\hat{\Phi}^2) + 2\hat{\Phi}\ \mathrm{Tr}(A_\mu \hat{\Phi}) \\ &= 4 A_\mu + 2 \hat{\Phi}\ \mathrm{Tr}(A_\mu \hat{\Phi}). \end{align}
If we define $a_\mu = -\frac{1}{2} \mathrm{Tr}(A_\mu \hat{\Phi})$ we get the expression for $A_\mu$ in the question. The derivation of $F_{\mu\nu}$ is very similar, so I won't write it out here unless someone asks.
As for $D_\mu \hat{\Phi} = 0$, note that the $U(1)$ unbroken by the Higgs is $U = \exp(i\alpha \hat{\Phi})$. To leave this unbroken, $F_{\mu\nu}$ must also be proportional to $\hat{\Phi}$. The Bogomol'nyi equation, $\vec{D}\Phi=\vec{B}$, becomes
$$ \vec{\nabla}h\ \hat{\Phi} + h \ \vec{D}\hat{\Phi} = \vec{b} \ \hat{\Phi}. $$
If we trace this against $\hat{\Phi}$, we find that $\vec{\nabla}h = \vec{b}$. Substituting back in gives $\vec{D}\hat{\Phi} = 0$.
Edit: In the non-BPS case we can apply the same tricks to the equation of motion for $F_{\mu\nu}$ to find $h^2 [D_\mu \hat{\Phi}, \hat{\Phi}] \propto f_\mu{}^\nu D_\nu \hat{\Phi}$. If we take the commutator with $\hat{\Phi}$ and trace against $D^\mu \hat{\Phi}$ we find $\lvert D\hat{\Phi} \rvert^2 = 0$. The time component vanishes for a static solution, so the space components do as well.