Theorem. A function $f : \mathbb{R}^{n_1} \times \mathbb{R}^{n_2} \to \mathbb{R}^{n_3}$ is bilinear iff each component of $f(v,w)$ is a linear combination of terms of the form $v_iw_j$, where $v=\begin{bmatrix}v_1&\cdots&v_{n_1}\end{bmatrix}^T$ and $w=\begin{bmatrix}w_1&\cdots&w_{n_2}\end{bmatrix}^T$ (i.e., terms from the outer product $v\otimes w$ of $v$ and $w$).
Proof ($\impliedby$). Follows immediately from the field axioms of $\mathbb{R}$.
Proof ($\implies$). This is the question. Or is the statement not even correct? I couldn't come up with a counterexample.
This is a sort of "follow up question" for this other question of mine: Simplest Examples for Tensors in Linear Algebra.
Does this theorem, if correct, generalize nicely to multilinear functions?
First write $$ v = \sum_{i=1}^{n_1} v_ie_i \quad \textrm{and} \quad w = \sum_{j=1}^{n_2} w_j \hat{e}_j $$ where $e_1,\dots,e_{n_1}$ is the standard basis for $\mathbb R^{n_1}$ and $\hat{e}_1,\dots,\hat{e}_{n_2}$ for $\mathbb R^{n_2}$. Now, since $f$ is bilinear, $$ f(u,v) = \sum_{i=1}^{n_1} v_if(e_i,w) = \sum_{i=1}^{n_1} \sum_{j=1}^{n_2} v_iw_j f(e_i,\hat{e}_j). $$