In order to discuss the equations of motion, it is advantageous to use eq. (\ref{eq:5}) to derive \begin{equation} \label{eq:7} d\xi^{i}\wedge d\xi^{j}\ =\ n\ \varepsilon^{ijk}\xi^{k}\ \omega_{(2)} \;\; , \;\; \varepsilon_{ijk}\ \xi^{i}\ d\xi^{j}\wedge d\xi^{k}\; =\; 2n\ \omega_{(2)} \; , \end{equation} where $\omega_{(2)}=\sin (\theta )d\theta\wedge d\varphi$ is the volume form of the unit-radius round 2-sphere; the field strength of the connection $\mathtt{A}$ then becomes \begin{equation} \label{eq:9} \mathtt{F}\ \equiv\ d\mathtt{A}\ +\ \mathtt{A}\wedge\mathtt{A} \ =\ \textstyle{1\over 4}\ dH^{-1}\wedge dH \ =\ \textstyle{n\over 2}\ \omega_{(2)}\ i\vec{\xi}\cdot\vec{\sigma} \; . \end{equation} Letting the generators of $\mathfrak{su}(2)$ be $\mathrm{T}_{a} = -\textstyle{i\over 2}\sigma^{a}$ such that $\left[\mathrm{T}_{a},\mathrm{T}_{b}\right]= \varepsilon_{abc}\mathrm{T}_{c}$, we see that \begin{equation} \label{eq:10} \mathtt{F}^{a} \; =\; -n\ \omega_{(2)}\ \xi^{a} \; . \end{equation} In order to treat the YM equations of motion it is fortuitous to write the $\mathfrak{su}(2)$-valued field strength as $\mathtt{F} = \aleph\ \omega_{(2)}\ H^{-1}$, where $\aleph$ is some purely imaginary constant which as far as the eoms are concerned is irrelevant; then we have \begin{eqnarray} \label{eq:11} 0 & =& \mathtt{D}\star\mathtt{F} \ \equiv\ d\star F \ +\ \mathtt{A}\wedge\star\mathtt{F} \ -\ \star \mathtt{F}\wedge \mathtt{A} \nonumber \\ & \simeq& d\left( H^{-1}\star\omega_{(2)}\right) \ +\ \textstyle{1\over 2} H^{-1}dH\wedge H^{-1}\ \star\omega_{(2)} \ -\ \textstyle{1\over 2}\star \omega_{(2)}\wedge H^{-1} H^{-1}dH \nonumber \\ & =& H^{-1}d\star\omega_{(2)} \; , \end{eqnarray} where in the last step we used the fact that $H$ is unitary and Hermitean, whence idempotent, i.e. $H^{2}=1$.
In conclusion, we need to show that $\star\omega_{(2)}$ is a closed 2-form. Since this depends on the metric, consider ordinary Minkowski space, the case of $\mathbb{R}^{3}$ being analogous, in spherical coordinates \begin{equation} \label{eq:12} ds^{2}\ =\ dt^{2} - dr^{2} - r^{2}d\theta^{2} - r^{2}\sin^{2}(\theta )d\varphi^{2} \hspace{.4cm}\mbox{whence}\hspace{.4cm} \star\omega_{(2)}\; =\; r^{-2}\ dr\wedge dt \; , \end{equation} which is clearly closed.
We are then ready to start combing the hedgehog: consider a transformation $U$ such that 3 $UHU^{-1}=\sigma^{3}$, i.e. a transformation that combs the hedgehog; explicitly this transformation is given by (See e.g. sec. 3.3.6 of ref. [3]) \begin{equation} \label{eq:4} \mathrm{SU}(2)\ni U\ =\ \textstyle{1\over \sqrt{2 (1+\xi_{3})}}\left( 1+\xi_{3}\ +\ i\ \left( \vec{e}_{3}\times\vec{\xi}\right)\cdot\vec{\sigma} \right) \; , \end{equation} which is clearly singular in the direction $\xi_{3}=-1$.
If we then make a gauge transformation on the gauge field then we find that \begin{equation} \label{eq:6} \mathtt{A}^{\prime} \ \equiv\ U\mathtt{A}U^{-1}\ -\ dU\ U^{-1} \; , \end{equation} which making use of eq. (\ref{eq:8}), leads to \begin{eqnarray} \label{eq:19} \mathtt{A}^{\prime}& =& \textstyle{1\over 2} UH^{-1}dH\ U^{-1}\ -\ dU\ U^{-1} \ =\ \textstyle{1\over 2} UH^{-1} d\left( HU^{-1}\right) -\textstyle{1\over 2} UdU^{-1} \ -\ dU\ U^{-1} \nonumber \\ & =& \textstyle{1\over 2} \sigma^{3} U\ d\left( U^{-1}\sigma^{3}\right) \ -\ \textstyle{1\over 2} dU\ U^{-1} \nonumber \\ & =& -\textstyle{1\over 2}\left( \sigma^{3}dU\ U^{-1}\sigma^{3}\ +\ dU\ U^{-1}\ \right) \; . \end{eqnarray} It is easy to see that the last equation is equal to \begin{equation} \label{eq:20} \mathtt{A}^{\prime}\; =\; -\pi_{3}\circ \left( dU\ U^{-1}\right) \; , \end{equation} where the notation $\pi_{3}$ means projection onto the $\sigma^{3}$ direction. This small result makes the verification of the relation between the Wu-Yang monopole and the Dirac monopole short and painless.
Indeed, using eq. (\ref{eq:5}) then gives that the explicit form reads \begin{equation} \label{eq:21} (\mathtt{A}^{\prime})^{3}\; =\; -\frac{\xi_{1}d\xi_{2}-\xi_{2}d\xi_{1}}{1+\xi_{3}}\; =\; -n\ \left( 1-\cos (\theta )\right)\ d\varphi \; . \end{equation} Decomposing it w.r.t. the natural Vielbein, we see that \begin{equation} \label{eq:22} \left(\mathtt{A}^{\prime}\right)^{3}_{3}\; =\; -\textstyle{n\over r}\ \frac{1-\cos (\theta )}{\sin (\theta )} \end{equation} from where the Dirac string is obvious.