= Solution
Use the orientation selected by the <volume form>, and let $\operatorname{vol}_g$ be its normalized <Riemannian volume form>. The <Euclidean metric> induces an <inner product> on the <exterior algebra> of covectors. The <Hodge star operator> is the unique map from $p$-forms to $(4-p)$-forms satisfying
$$
\alpha\wedge *\beta=\langle\alpha,\beta\rangle_g\operatorname{vol}_g.
$$
For an oriented orthonormal coframe $e^1,\ldots,e^4$, it sends a basis wedge to the complementary wedge with the sign of the permutation that restores $e^1\wedge\cdots\wedge e^4$. Applying the <Hodge star operator> twice exchanges blocks of $p$ and $4-p$ covectors, so
$$
*^2=(-1)^{p(4-p)},\qquad\boxed{*^2=1\quad\text{on two-forms}.}
$$
Here the supplied volume is interpreted in the usual metric-normalized sense. If instead one defines the operator with an arbitrary unnormalized $\operatorname{vol}=f\operatorname{vol}_g$, that operator is $f*_g$ and its square on two-forms is $f^2$. The usual self-duality statements use the metric-normalized <Hodge star operator>, with the supplied volume specifying orientation.
The projections $P_\pm=(1\pm*)/2$ give the <Hodge splitting of Euclidean two-forms>:
$$
\boxed{\Lambda^2=\Lambda^2_+\oplus\Lambda^2_-,\qquad
\alpha_\pm=\frac12(\alpha\pm*\alpha),\qquad *\alpha_\pm=\pm\alpha_\pm.}
$$
Both spaces have dimension three. With $e^4=dt$, bases are
$$
\Sigma_i^\pm=e^i\wedge dt\ \pm\ \frac12\epsilon_{ijk}e^j\wedge e^k,
\qquad i=1,2,3.
$$
The <Hodge star operator> is an orthogonal involution on two-forms and therefore is self-adjoint. For a <self-dual two-form> $H$ and an <anti-self-dual two-form> $G$,
$$
\langle H,G\rangle=\langle *H,G\rangle
=\langle H,*G\rangle=-\langle H,G\rangle=0.
$$
Consequently
$$
\boxed{H\wedge G=\langle H,*G\rangle\operatorname{vol}_g=0.}
$$
This is the <wedge orthogonality of opposite-duality two-forms>.
For the <Yang-Mills action>, take an anti-Hermitian <special unitary group> connection and the fundamental matrix <trace>, so the positive invariant pairing on its <Lie algebra> is $\langle X,Y\rangle=-\operatorname{tr}(XY)$. Write its <gauge curvature> as $F=F_++F_-$ using the <Hodge splitting of Euclidean two-forms>, and define
$$
\|F_\pm\|^2=-\int\operatorname{tr}(F_\pm\wedge *F_\pm),\qquad
S_{\mathrm{YM}}=-\frac1{g_{\mathrm{YM}}^2}\int\operatorname{tr}(F\wedge*F).
$$
Cross terms vanish by the <wedge orthogonality of opposite-duality two-forms>. Thus
$$
S_{\mathrm{YM}}=\frac{\|F_+\|^2+\|F_-\|^2}{g_{\mathrm{YM}}^2},\qquad
\int\operatorname{tr}(F\wedge F)=-\|F_+\|^2+\|F_-\|^2.
$$
With this anti-Hermitian convention the <Second Chern number> is
$$
k=c_2(E)[S^4]=\frac1{8\pi^2}\int\operatorname{tr}(F\wedge F).
$$
For example, this normalization follows by expanding $\det(1+iF/(2\pi))$ and using $\operatorname{tr}F=0$. Interpreting the integral as an integer <Second Chern number> on $\mathbb R^4$ assumes the usual decay and gauge behavior that allow extension over the point at infinity. The following norm inequality itself does not require integrality:
$$
\boxed{S_{\mathrm{YM}}\ge\frac{8\pi^2}{g_{\mathrm{YM}}^2}|k|.}
$$
The <Yang-Mills instanton Bogomolny bound> is saturated precisely when one component vanishes: $F=*F$ or $F=-*F$. In the stated convention the self-dual case has $k\le0$ and the anti-self-dual case has $k\ge0$. Reversing orientation, or defining topological charge with the opposite sign, reverses this assignment while leaving the absolute-value bound unchanged.
For the <self-dual Yang-Mills equations in temporal gauge>, choose $\operatorname{vol}_g=dx^1\wedge dx^2\wedge dx^3\wedge dt$ and $\epsilon_{123}=1$. The relevant <Hodge star operator> identities are
$$
*(dt\wedge dx^i)=-\frac12\epsilon_{ijk}dx^j\wedge dx^k,
\qquad *(dx^j\wedge dx^k)=-\epsilon_{ijk}dt\wedge dx^i.
$$
Writing $F=F_{ti}\,dt\wedge dx^i+\tfrac12F_{jk}\,dx^j\wedge dx^k$, the <self-dual Yang-Mills equations> $F=*F$ give
$$
F_{ti}=-\frac12\epsilon_{ijk}F_{jk}.
$$
In <temporal gauge>, $A_t=0$, so the <gauge curvature> component is $F_{ti}=\partial_tA_i-\partial_iA_t+[A_t,A_i]=\partial_tA_i$. Hence
$$
\boxed{\partial_tA_i=-\frac12\epsilon_{ijk}F_{jk},\qquad
F_{jk}=\partial_jA_k-\partial_kA_j+[A_j,A_k].}
$$
The minus sign follows from placing $dt$ last in the orientation and first in the mixed curvature component.
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