Let's say $V$ and $W$ are two finite dimensional vector spaces over $\mathbb R$. A bilinear form $\mathrm B : V \times W \to \mathbb R$ is necessarily linear in both factors: $\mathrm B(v_1+v_2,w) = \mathrm B(v_1,w)+\mathrm B(v_2,w)$ for all $v_i \in V$ and $w \in W$; $\mathrm B(v,w_1+w_2) = \mathrm B(v,w_1)+\mathrm B(v,w_2)$ for all $v \in V$ and $w_j \in W$; and $\mathrm B(\lambda v,w) = \lambda \mathrm B(v,w)=\mathrm B(v,\lambda w)$ for all $\lambda \in \mathbb R$, $v \in V$ and $w \in W$.
The restriction of $\mathrm B$ gives elements of the duals $V'$ and $W'$.
For a fixed $v_0 \in V$, we get a linear map $W \to \mathbb R$ given by $w \mapsto \mathrm B(v_0,w)$. Hence, $v \mapsto (w \mapsto \mathrm B(v,w))$ is a linear map $V \to W'$. Similarly, for a fixed $w_0 \in W$, we get a linear map $V \to \mathbb R$ given by $v \mapsto \mathrm B(v,w_0)$. Hence $w \mapsto \left( v \mapsto \mathrm B(v,w) \right)$ is a linear map $W \to V'$.
What can be said about the mapping $V \times W \to W' \times V'$, where $$(v,w) \mapsto \left(w \mapsto \mathrm B(v,w), v \mapsto \mathrm B(v,w)\right)$$
I got this construction from thinking about $\mathrm B : V \times V' \to \mathbb R$ given by $\mathrm B(v,\varphi) := \varphi(v)$, and the map $V \to \left(V'\right)'$ given by $v \mapsto \left(\varphi \mapsto \varphi(v)\right)$ shows that $V \cong \left(V'\right)'$.