Given there a vector space $V$ with a scalar product $g(v_1,v_2)$ on it, what is the scalar product on, say, $V \otimes V^*$ ?
According to Jeffrey Lee's "Manifolds and Differential Geometry" (see 7.6 Metric Tensors):
...there is a unique scalar product $g^1_1$ on $V \otimes V^*$ such that for $v_1 \otimes \alpha_1$ and $v_2 \otimes \alpha_2$ $\in V \otimes V^*$ we have $$ g_1^1 \left(v_1 \otimes \alpha_1), (v_2 \otimes \alpha_2) \right) = g(v_1,v_2) \, g^*(\alpha_1, \alpha_2)$$
Here $g^*$ is a scalar product on $V^*$: $$g^*(\alpha, \beta) = g(\alpha^\sharp, \beta^\sharp)$$
What bothers me is that I do not understand why it couldn't be $$ g_1^1 \left(v_1 \otimes \alpha_1), (v_2 \otimes \alpha_2) \right) = g(v_1,v_2) + g^*(\alpha_1, \alpha_2)$$
Is there a reason to choose the first definition? Particularly given how metric tensor is defined of products of Riemannian manifolds.
For $V \otimes V$ the book earlier defines scalar product as: $$ g \left(v_1 \otimes u_1), (v_2 \otimes u_2) \right) = g(v_1,v_2) \, g(u_1, u_2)$$
The map $$g_1^1 \left(v_1 \otimes \alpha_1), (v_2 \otimes \alpha_2) \right) = g(v_1,v_2) + g^*(\alpha_1, \alpha_2)$$ isn't even well-defined: For all $\lambda \in \mathbb{R} - \{0\}$ we have $$v_1 \otimes \alpha_1 = (\lambda^{-1} v_1) \otimes (\lambda \alpha_1),$$ but $$g(\lambda^{-1} v_1,v_2) + g^*(\lambda \alpha_1, \alpha_2) = \lambda^{-1} g(v_1,v_2) + \lambda g^*(\alpha_1, \alpha_2),$$ which in general does not coincide with $$g(v_1,v_2) + g^*(\alpha_1, \alpha_2).$$
What you probably have in mind is this: Given scalar products $g, h$ respectively on vector spaces $\mathbb{V}, \mathbb{W}$ we get a natural scalar product on the direct sum $\mathbb{V} \oplus \mathbb{W}$ defined by $$((v_1, w_1), (v_2, w_2)) \mapsto g(v_1, v_2) + h(w_1, w_2)).$$