How do we define the Lie bracket on the complexification $\mathfrak{g} \otimes_\Bbb{R} \Bbb{C}$?

First of all, it seems the right extension is the following: $$[v\otimes\lambda,w\otimes\mu]:=[v,w]\otimes\lambda\mu.$$

This satisfies bilinearity, and Jacobi's identity. However, how can we show that this is the unique extension of the Lie bracket? We have the following result (taken, for example, from Bump's Lie groups):

Proposition: If $V$ and $U$ are real vector spaces, any $\mathbb R$-bilinear map $V\times V\to U$ extends uniquely to a $\mathbb C$-bilinear map $V_{\mathbb C}\times V_{\mathbb C}\to U_{\mathbb C}$.

Proof: This basically follows from the properties of tensor products. Any $\mathbb R$-bilinear map $V\times V\to U$ corresponds to a unique $\mathbb R$-linear map $V\otimes_{\mathbb R} V\to U$. But any $\mathbb R$-linear map extends uniquely to a $\mathbb C$-linear map of the complexified vector spaces (this is easy to prove). Hence, we have a $\mathbb C$-linear map $(V\otimes_{\mathbb R} V)_{\mathbb C}\to U_{\mathbb C}$. But we have the following isomorphism: $$(V\otimes_{\mathbb R} V)_{\mathbb C}\cong V_{\mathbb C}\otimes_{\mathbb C} V_{\mathbb C};$$ on the left-hand side, the tensor product is over $\mathbb R$, and on the right-hand side, it is over $\mathbb C$. Finally, our $\mathbb C$-linear map $V_{\mathbb C}\otimes_{\mathbb C} V_{\mathbb C}\to U_{\mathbb C}$ corresponds to a unique $\mathbb C$-bilinear map $V_{\mathbb C}\times V_{\mathbb C}\to U_{\mathbb C}$.

Using the suggestion of MTurgeon we check that $g$ is a compatible with the Lie bracket on the complexification:

$$\begin{eqnarray*} g[v\otimes \lambda, w\otimes \mu] &=& g([v,w]_\mathfrak{g}\otimes \lambda\mu)\\ &=&\lambda\mu g([v,w]_\mathfrak{g} \otimes 1)\\ &=&\lambda \mu h([v,w]_\mathfrak{g}) \\ &=&\lambda\mu \big[h(v),h(w)\big]_\mathfrak{h}\\ &=& \big[ \lambda h(v),\mu h(w)\big]_\mathfrak{h}\\ &=&\bigg[g(v \otimes \lambda),g(w \otimes \mu) \bigg]_\mathfrak{h}. \end{eqnarray*}$$

Edit: For those curious about MTurgeon's isomorphism below, recall this is coming from a more general result concerning isomorphisms involving extensions of scalars. The relevant isomorphism I will put here is Theorem 6.15 (2) in here.

Let $M$ be an $R$ - module and $N$ and $S$ - module with $f : R\to S$ a ring homomorphism. Then the $S$ - module $M \otimes_R N$ is isomorphic to $(S \otimes_R M)\otimes_S N$ by sending $1 \otimes m$ to $(1 \otimes m) \otimes n$.

MTurgeon's isomorphism now falls out applying $R= \Bbb{R}$, $S = \Bbb{C}$, $M = V$ and $ N = V \otimes_\Bbb{R} \Bbb{C}$. Recall that $N$ is a $\Bbb{C}$ - module by extension of scalars with complex multiplication defined on elementary tensors as

$$\alpha( v \otimes \beta) = v \otimes (\alpha\beta)$$

for all $\alpha,\beta \in \Bbb{C}$ and $v \in V$.

Another guess would be this?

$[v \otimes i,w \otimes i] = [v,w] \otimes i$

or $[v \otimes i,w \otimes i] = [v,w] \otimes -1$?

I'm not quick enough to verify. In any case it seems you'd want to confine the defined bracket's working parts to the Lie algebra, and avoid involving the commutative ring.