OK. So let $\frak{g}$ be a real Lie algebra. It's complexification ${\frak{g}} \otimes_\mathbb{R} \mathbb{C}$ can be identified with $\frak{g} \oplus \frak{g}$ in an obvious way.
Now, the Killing form is $B(\cdot, \ast) = \operatorname{tr}(\operatorname{ad}_\cdot \circ \operatorname{ad}_\ast)$. Given $v, w \in \frak{g}$, we want to compare $B_{\frak{g}}(v, w)$ and $B_{\frak{g} \oplus \frak{g}}((v, 0), (w, 0))$.
Assume $(x, y) \in \frak{g} \oplus \frak{g}$, then an easy computation shows that
\begin{eqnarray*} \operatorname{ad}_{(v, 0)} \circ \operatorname{ad}_{(w, 0)} ((x, y)) &=& [(v, 0), [(w, 0), (x, y)]_{\frak{g} \oplus \frak{g}}]_{\frak{g} \oplus \frak{g}} \\ &=& ([v, [w, x]_{\frak{g}}]_{\frak{g}}, [v, [w, y]_\frak{g}]_\frak{g}) \end{eqnarray*}
So $$\operatorname{ad}_{(v, 0)} \circ \operatorname{ad}_{(w, 0)} = \operatorname{ad}_{v} \circ \operatorname{ad}_{w} \otimes \operatorname{ad}_{v} \circ \operatorname{ad}_{w},$$ would not that imply that $B_{\frak g \oplus g} = 2 B_\frak{g}$? This is not correct, so what is wrong?
The Killing form $B_{\mathfrak{g}}$ of $\mathfrak{g}$ is a real bilinear form, while the Killing form $B_{\mathfrak{g}_{\mathbb{C}}}$ of $\mathfrak{g}_{\mathbb{C}}:=\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ is a complex bilinear form, so they cannot be proportional in any way. However, we have $$B_{\mathfrak{g}_{\mathbb{C}}}|_{\mathfrak{g}\times \mathfrak{g}}=B_{\mathfrak{g}}$$ which is probably the formula you are looking for.
Similarly, if we denote by $(\mathfrak{g}_{\mathbb{C}})^{\mathbb{R}}$ the realification of $\mathfrak{g}_{\mathbb{C}}$, i.e. the underlying real Lie algebra of $\mathfrak{g}_{\mathbb{C}}$ of twice the dimension, then $$B_{(\mathfrak{g}_{\mathbb{C}})^{\mathbb{R}}}=2\operatorname{Re}B_{\mathfrak{g}_{\mathbb{C}}}.$$ See Knapp Lie groups Beyond an Introduction p.58.