Let $K$ be a compact, connected, simply connected Lie group with Lie algebra $\mathfrak k$ and Killing from $B_{\mathfrak k}$. It is well known that $B_{\mathfrak k}$ is a negative definite symmetric bilinear form on $\mathfrak k$ and hence (its negative) defines an inner product. We can extend the Killing form to a Hermitian inner product on $\mathfrak k_{\mathbb C}$, which I will denote by $B_{\mathfrak k}^{\mathbb C}$.
Now the complex semi-simple Lie algebra $\mathfrak k_{\mathbb C}$ itself comes equipped with a Killing form $B_{\mathfrak k_{\mathbb C}}$. This form is symmetric and bilinear, but is not negative definite. Indeed, if $$ \mathfrak k_{\mathbb c} = \mathfrak h \oplus \bigoplus_{\alpha \in R} \mathfrak g_\alpha$$ is a root-space decomposition of $\mathfrak k_{\mathbb C}$ then one immediately has that $B_{\mathfrak k_{\mathbb C}}(\mathfrak g_\alpha, \mathfrak g_{\beta}) = 0$ whenever $\alpha + \beta \neq 0$.
I'm looking for a relationship between $B_{\mathfrak k}^{\mathbb C}$ and $B_{\mathfrak k_{\mathbb C}}$.
I would posit a guess that $B_{\mathfrak k}^{\mathbb C}(X,Y) = B_{\mathfrak k_{\mathbb C}}(X,\bar Y)$ where $\bar Y$ is complex conjugation. In particular, we know that $B_{\mathfrak k_{\mathbb C}}$ is bilinear, so composition with conjugation in the second coordinate would result in conjugate linearity in the second coordinate, much akin to the Hermitian inner product. Furthermore, we know that $\overline{\mathfrak g_\alpha}= \mathfrak g_{-\alpha}$ which would also nicely tie the two together.
I am not certain how to turn this collection of corroborating facts into a proof. If anyone has any insight it would be much appreciated.
I don't know to much about root-space decompositions, but does this have to do with the Cartan decoposition:
If we have a real semi-simple Lie algebra $\mathfrak{g}$ and let $\mathfrak{g}_{\mathbb{C}}=\mathbb{C}\otimes_{\mathbb{R}}\mathfrak{g}$. A real form is any real subalgebra $\mathfrak{g}'$ such that $\mathbb{C}\otimes_{\mathbb{R}}\mathfrak{g}'=\mathfrak{g}_{\mathbb{C}}$. Then it is a theory of Weyl that there is a real form $\mathfrak{u}$ where the such that the killing form is neg-def on $\mathfrak{u}$. You can eventually show that $$\mathfrak{g}=(\mathfrak{g}\cap\mathfrak{u})\oplus (\mathfrak{g}\cap i\mathfrak{u})$$ which is called the Cartan decomposition and typically we write $\mathfrak{k}=\frak{g}\cap\frak{u}$ and $\mathfrak{p}=\mathfrak{g}\cap i\mathfrak{u}$.
The reason I bring this up is because along the way we fixed an involution $\tau$ (which we later showed was conjugate-linear) of $\frak u$ and defined $(,)$ a neg-def Hermitian product on $\mathfrak{g}_{\mathbb{C}}$ as $$(X,Y)=-B_{\mathfrak{g}_{\mathbb{C}}}(X,\tau Y).$$
I think that this $(X,Y)$ is your $B_{\mathfrak{k}}^{\mathbb{C}}$ and my $B_{\mathfrak{g}_{\mathbb{C}}}$ is your $B_{\mathfrak{k}_{\mathbb{C}}}$. I also think my $\tau$ corresponds to your complex conjugation as complex conjugation is a conjugate-linear involution and there is a correspondence between real forms and involutions (I think).
I know this isnt really the problem you are discussing, but what I am saying maybe the construction of the Cartan Decomposition might give some insight. I hope you found that helpful.