Theorem: Let $U$ and $V$ be vector spaces. Let $\mathbf{u}^* \in U^*$. Define $\mathbf{f} : U \otimes V \to V$: $$\mathbf{f}\left(\sum_{r} \mathbf{u}_r \otimes \mathbf{v}_r\right) = \sum_{r} \mathbf{u}^*(\mathbf{u}_r) \mathbf{v}_r$$ Then $$\sum_{r} \mathbf{u}_r \otimes \mathbf{v}_r = \sum_{r} \mathbf{u}_r' \otimes \mathbf{v}_r' \Rightarrow \mathbf{f}\left(\sum_r \mathbf{u}_r \otimes \mathbf{v}_r\right) = \mathbf{f}\left(\sum_{r} \mathbf{u}_r' \otimes \mathbf{v}_r'\right)$$
How to prove this theorem? I tried to use the fact that $\otimes$ is bilinear: \begin{align*} \mathbf{f}\left(\sum_{r} \mathbf{u}_r \otimes \mathbf{v}_r\right) &= \sum_{r} \mathbf{f}(\mathbf{u}_r \otimes \mathbf{v}_r) \\ &= \text{I don't know what would be here} \\ &= \sum_{r} \mathbf{f}(\mathbf{u}_r' \otimes \mathbf{v}_r') \\ &= \mathbf{f}\left(\sum_{r} \mathbf{u}_r' \otimes \mathbf{v}_r'\right) \end{align*}
just define $f$ on cartesian product and use universal theorem to conclude it.