This actually can be done in much greater generality.
Let $G$ be a compact group and $K \subseteq G$ a closed subgroup.
Then for any finite-dimensional representation $(V,\pi)$ of $K$ you can form the associated bundle $G \times_K V$ over $G/K$.
Sections of this bundle are given by functions $f : G \to V$ satisfying the equivariance condition
$$
f(xs) = \pi(s)^{-1} f(x)
$$
for all $x \in G$ and $s \in K$.
You are now asking for a collection of sections that generates the module of all sections of this bundle.
One way to do this is via Frobenius reciprocity.

Frobenius reciprocity implies that there is a representation $(W,\sigma)$ of $G$ such that $V \subseteq W$ and $\sigma(s)v = \pi(s)v$ for $v \in V$.
Since $G$ is compact, there is a $G$-invariant inner product on $W$.
Let $P$ be the orthogonal projection of $W$ onto $V$ with respect to this inner product.

Then choose an orthonormal basis $(w_i)$ for $W$.
For each $i$, define a function $\eta_i : G \to V$ by
$$
\eta_i(x) = P \sigma(x)^{-1}w_i.
$$
It is not too hard to show that each $\eta_i$ is actually a section of the associated bundle.

To show that these sections generate the module, define a $C^\infty(G/K)$-valued pairing on the module $\Gamma = \Gamma^\infty(G/K, G \times_K V)$ by
$$
\langle \zeta, \xi \rangle(x) = \langle \zeta(x), \xi(x) \rangle,
$$
for sections $\zeta, \xi$, where $\langle , \rangle$ was the invariant inner product on $W$ that we chose above.
Then it turns out that for any $\xi \in \Gamma$, we have
$$
\xi = \sum_i \langle \xi, \eta_i \rangle \eta_i.
$$
When verifying this you need to use the fact that since the inner product on $W$ is $G$-invariant, any element $x$ of $G$ takes the orthonormal basis $(w_i)$ to another orthonormal basis.

Anyway, this is only as canonical as choosing an orthonormal basis of $W$, so it may not be what you want. But it is at least a way of getting a nice generating set.

Edit: I should explain what this is really doing. Note that a homogeneous vector bundle bundle $G \times_K V$ over $G/K$ will be a trivial bundle if and only if the representation $(V,\pi)$ of $K$ is actually the restriction of a representation of $G$ on $V$. To see this, note that if this is the case, then in the construction above we can take $W = V$, the projection $P$ is just the identity operator, and the global sections $\eta_i$ vanish nowhere and form a global frame for the bundle.

In the case when we have to take $W$ to be strictly larger than $V$, what we are doing is embedding the homogeneous bundle $G \times_K V$ into the trivial bundle $G \times_K W$. The projection $P$ tells us how to cut down fiberwise from the trivial bundle to the nontrivial one.

I should add that in your case, you can take the larger vector space $W$ to be $\mathcal{G}^\ast$ itself with the coadjoint action of $G$, and you can take the invariant inner product to be the one induced by the Killing form on $\mathcal{G}$.