I'm trying to show that $\alpha G^a_{\mu\nu} \widetilde{G^a_{\mu\nu}}$ can be written as a total derivative, where $$\begin{equation} G^a_{\mu\nu} = \partial_{\mu} G^a_{\nu} - \partial_{\nu}G^a_{\mu}-gf_{bca}G^b_{\mu}G^c_{\nu} \end{equation} $$ $$\widetilde{G^a_{\mu\nu}} = \frac{1}{2} \epsilon^{\mu\nu\lambda\rho}G^a_{\lambda\rho} $$ And $\epsilon$ is the Levi-Civita symbol. Then expand $G^a_{\lambda\rho}$ using the first equation, multiply the two and expand the brackets: $$\frac{1}{2} \epsilon^{\mu\nu\lambda\rho} (\partial_{\lambda}G^a_{\rho}\partial_{\mu}G^a_{\nu}-\partial_{\lambda}G^a_{\rho}\partial_{\nu}G^a_{\mu}-\partial_{\lambda}G^a_{\rho}gf_{bca}G^b_{\mu}G^c_{\nu}-\partial_{rho}G^a_{\lambda}\partial_{\mu}G^a_{\nu}+\partial_{\rho}G^a_{\lambda}\partial_{\nu}G^a_{\mu}+\partial_{\rho}G^a_{\lambda}gf_{bca}G^b_{\mu}G^c_{\nu}-gf_{bca}G^b_{\lambda}G^c_{\rho}\partial_{\mu}G^a_{\nu}+gf_{bca}G^b_{\lambda}G^c_{\rho}\partial_{\nu}G^a_{\mu}+g^2f^2_{bca}G^b_{\lambda}G^c_{\rho}G^b_{\mu}G^c_{\nu})$$
So that looks a mess, but because the Levi-Civita symbol is antisymmetric that equals $$\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}(4\partial_{\lambda}G^a_{\rho}\partial_{\mu}G^a_{\nu}+2\partial_{\rho}G^a_{\lambda}gf_{bca}G^b_{\mu}G^c_{\nu}+2gf_{bca}G^b_{\lambda}G^c_{\rho}\partial_{\nu}G^a_{\mu}+g^2f^2_{bca}G^b_{\lambda}G^c_{\rho}G^b_{\mu}G^c_{\nu})$$
What it's meant to be, somehow, is $$\partial_{\mu}\left(\alpha \epsilon^{\mu\nu\lambda\rho}G^a_{\nu}(G^a_{\lambda\rho} +\frac{1}{3}g f_{bca}G^b_{\lambda}G^c_{\rho}) \right)$$
But I really, really can't see how you'd get there!
$\newcommand{\d}{\mathrm{d}}$ $\newcommand{\i}{\mathrm{i}}$ $\newcommand{\Tr}{\mathrm{Tr}\,}$
Switch to math conventions (who can deal with all those factors of $\i$ and $g$?): $$\begin{align} A^a_{\mu}&=-\i g G^a_{\mu}\\ F^a_{\mu\nu}&=-\i g G^a_{\mu\nu}\\ \tau_a&=\i T_a \\ c^a_{bc}&=-f^a_{bc}\\ [\tau_a,\tau_b]&=c^c_{ab}\tau_c\text{.} \end{align}$$
Let $$\kappa_{ab}=\langle \tau_a,\tau_b\rangle \propto \Tr \tau_a\tau_b$$ be the components of the Killing form in negative-definite convention.
I freely use the notation of Lie algebra-valued forms: $$\begin{align}\langle \alpha \wedge \beta \rangle&=\langle \tau_a,\tau_b\rangle \alpha^a\wedge \beta^b=\kappa_{ab}\alpha^a\wedge\beta^b \\ [\alpha\wedge\beta]&=[\tau_a,\tau_b]\alpha^a\wedge\beta^b=\tau_cc^c_{ab}\alpha^a\wedge\beta^b\text{.} \end{align}$$
With these conventions, $$\begin{align}F&=\d A +\tfrac{1}{2}[A\wedge A]\\ F^a_{\mu\nu}&=2\partial_{[\mu}A^a_{\nu]}+c^a_{bc}A^{b}_{[\mu}A^c_{\nu]} \end{align}$$
Define the Chern–Weil $4$-form $W$ $$\begin{align} W&=\langle F\wedge F\rangle \\ W_{\mu\nu\rho\sigma}&=6\kappa_{ab} F_{[\mu\nu}^a F_{\rho\sigma]}^b \end{align}$$ and the Chern–Simons $3$-form $S$ $$\begin{align}S&=\langle A\wedge F\rangle -\tfrac{1}{6}\langle A\wedge [A\wedge A]\rangle \\ S_{\nu\rho\sigma}&=3\kappa_{ab}A^a_{[\nu} F^b_{\rho\sigma]} -\kappa_{ab}c^{b}_{cd}A_{[\nu}^a A_{\rho}^c A_{\sigma]}^d\text{.} \end{align}$$
Then the task is to show $$\begin{align} W&=\d S\\ W_{\mu\nu\rho\sigma}&=4\partial_{[\mu}S_{\nu\rho\sigma]} \end{align}$$
Differentiating the individual terms of $S$ gives $$\begin{split}\d\langle A\wedge F\rangle &=\langle \d A\wedge F\rangle -\langle A \wedge \d F\rangle\\ &=\langle (F-\tfrac{1}{2}[A\wedge A])\wedge F\rangle +\langle A \wedge [A\wedge F]\rangle\\ &=(\langle F\wedge F\rangle-\tfrac{1}{2}\langle [A\wedge A]\wedge F\rangle) +\langle [A \wedge A]\wedge F\rangle\\ &=\langle F\wedge F\rangle+\tfrac{1}{2}\langle [A\wedge A]\wedge F\rangle \end{split}$$
$$\begin{split}\d\langle A\wedge [A\wedge A]\rangle &=\langle \d A\wedge [A\wedge A]\rangle -\langle A\wedge [\d A\wedge A]\rangle + \langle A\wedge [A\wedge \d A]\rangle\\ &=3\langle [A\wedge A]\wedge \d A\rangle \\ &=3\left(\langle [A\wedge A]\wedge F\rangle -\tfrac{1}{2} \langle [A\wedge A]\wedge [A\wedge A]\rangle\right)\\ &=3\langle [A\wedge A]\wedge F\rangle\text{.} \end{split}$$
The sum is $$\begin{split}\d S &=\d\langle A\wedge F\rangle -\tfrac{1}{6}\d \langle A\wedge [A\wedge A]\rangle\\ &=(\langle F\wedge F\rangle+\tfrac{1}{2}\langle [A\wedge A]\wedge F\rangle)-\tfrac{1}{2}\langle [A\wedge A]\wedge F\rangle\\ &=\langle F \wedge F\rangle \\ &= W\text{.} \end{split}$$
(This proof is an adaptation of Freed (1992), Proposition 1.27.)
Aside from the usual calculus of differential forms (or Ricci calculus of antisymmetric tensors, if you want to use index notation), the important ingredients are the Jacobi identity
$$\begin{align}[\alpha\wedge[\beta\wedge\gamma]]&=[[\alpha\wedge\beta]\wedge\gamma]+(-1)^{\lvert\alpha\rvert\lvert\beta\rvert}[\beta\wedge[\alpha\wedge\gamma]]\\ c^a_{ub}c^b_{vw}&=c^a_{bw}c^{b}_{uv}+c^a_{vb}c^b_{aw}\text{,} \end{align}$$
invariance of the Killing form
$$\begin{align} \langle [\alpha\wedge \beta]\wedge \gamma \rangle +(-1)^{\lvert\alpha\rvert \lvert \beta \rvert}\langle\beta \wedge[\alpha\wedge\gamma]\rangle &=0\\ \kappa_{aw}c^a_{uv}+\kappa_{va}c^a_{uw}&=0\text{,} \end{align}$$
and the second Bianchi identity $$\begin{align}\d F+[A\wedge F]&=0 \\ 3\partial_{[\nu}F^a_{\rho\sigma]}+3c^a_{bc}A^b_{[\nu}F^c_{\rho\sigma]} &=0\text{.} \end{align}$$