I'm trying to understand a step in a proof:
Let $\mathfrak{g} = \mathfrak{so}(3,2) \simeq \mathfrak{sp}(2, \mathbb{R})$ with the root decomposition $\mathfrak{g} = \underbrace{\mathfrak{g}^{-2} + \mathfrak{g}^{-1}}_{:=\mathfrak{g}_-} + \mathfrak{h} + \underbrace{\mathfrak{g}^{1} + \mathfrak{g}^{2}}_{:=\mathfrak{g}_+}$
In this setting, the author of the proof says via the Cartan-Killing form of $\mathfrak{g}$, the space $\text{Hom}_k(\mathfrak{g}_-, \mathfrak{g})$ can be identified with $\text{Hom}_k(\mathfrak{g}_+, \mathfrak{g}^*)$ where $\text{Hom}_k(\mathfrak{g}_-, \mathfrak{g}):= \{A \in \text{Hom}(\mathfrak{g}_-, \mathfrak{g}) \enspace | \enspace A(\mathfrak{g}^i) \subset \mathfrak{g}^{i+k}\}$
Why is this identification possible?
I know that $\kappa$ the Cartan-Killing form of $\mathfrak{g}$ is non-degenerate on $\mathfrak{g}$ and that $\kappa$ induces a nondegenerate pairing of $\mathfrak{g}^{\alpha}$ and $\mathfrak{g}^{-\alpha}$ (Proposition 6.3.2 in Hilgert and Neeb's Structure and Geometry of Lie Groups (2012))
I've got that $$\text{Hom}_(\mathfrak{g}_-, \mathfrak{g}) \simeq (g_{-})^* \otimes \mathfrak{g}$$ and via the non-degeneracy of $\kappa|_{\mathfrak{g}^{\alpha, -\alpha}}$, have an induced isomorphism $$\mathfrak{g}^{\alpha} \to (\mathfrak{g}^{-\alpha})^*,$$ and so have $$(\mathfrak{g}_-)^* \otimes \mathfrak{g} \simeq \mathfrak{g}_+ \otimes \mathfrak{g}.$$ There is also the induced isomorphism $$\mathfrak{g} \to \mathfrak{g}^*$$ via the non-degeneracy of $\kappa$ on $\mathfrak{g}$ yielding $$\mathfrak{g}_+ \otimes \mathfrak{g}^*.$$
I get stuck here and feel like I'm missing something really obvious but am not sure what to consider.
So first things first, that is not a "root decomposition" since $\mathfrak{g}^1$ etc. are not root spaces. Instead I would call it a Lie algebra grading. Secondly, I will also note there are differences of notation around the symplectic Lie algebras. I would usually use $\mathfrak{sp}(2, \mathbb{R})$ to refer to the Lie algebra isomorphic to $\mathfrak{sl}(2, \mathbb{R}) \cong \mathfrak{so}(2, 1)$ rather than $\mathfrak{so}(3, 2)$ which I would identify with $\mathfrak{sp}(4, \mathbb{R})$.
On to the main question. As you have found there is a canonical identification of $\mathrm{Hom}(\mathfrak{g}_{-}, \mathfrak{g})$ with $\mathfrak{g}_{+} \otimes \mathfrak{g}$ which we could identify by swapping the order of terms with $\mathfrak{g}\otimes \mathfrak{g}_{+} = \mathrm{Hom}(\mathfrak{g}^*,\mathfrak{g}_{+})$ but not the other way round. I don't believe there is a natural identification as stated. You don't define $\mathrm{Hom_k}(\mathfrak{g}_{+},\mathfrak{g}^*)$ but I can't see an identification any which way round without adding in some other structure.