The trace is simply a (properly normalised) ad-invariant inner product on the Lie algebra; that is, a nondegenerate symmetric bilinear form $\langle-,-\rangle$ which obeys the "associativity" condition $$\langle [x,y],z \rangle = \langle x, [y,z] \rangle$$ for every $x,y,z$ in $\mathfrak{g}$.
Lie algebras admitting such inner products are said to be metric. The normalisation of the inner product is such that $k$ is an integer. This only makes sense for indecomposable metric Lie algebras; that is, those which are not isomorphic to the direct product of perpendicular proper ideals.
The notation "tr" stems from the fact that if $\rho: \mathfrak{g} \to \operatorname{End}(V)$ is a faithful finite-dimensional representation, then $$\langle x, y\rangle := c \operatorname{tr}\rho(x)\rho(y)$$ works for a suitable nonzero $c$. (For a simple Lie algebra, just take $\rho$ to be the adjoint representation.)
For the explicit case of $\mathfrak{g}$ the Lie algebra of SU(2) you can take $\rho$ to be the fundamental representation and $c= -\frac12$, I believe.
Edit
Notice that $\operatorname{tr}(A \wedge dA)$ is really $\langle A \stackrel{\wedge}{,} dA \rangle$, where $\langle -\stackrel{\wedge}{,}-\rangle$ means that we are both taking the wedge product of the forms and the inner product on the Lie algebra. Similarly, $$\operatorname{tr}(A \wedge A \wedge A) = \frac12 \langle [A\stackrel{\wedge}{,}A] \stackrel{\wedge}{,} A \rangle,$$ with a similar notational caveat about $[A\stackrel{\wedge}{,}A]$.
Further addition
In response to Anirbit's comment, I would say that there is, in general, no trace on vector-valued differential forms; although if the forms take values in endomorphisms, then of course there is: simply compose with the trace of endomorphisms to obtain a map $$\Omega^\bullet(M;\operatorname{End}(V)) \to \Omega^\bullet(M).$$
