What is the scalar product of tensors? 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:
$$ g \left(v_1 \otimes u_1),  (v_2 \otimes u_2) \right) = g(v_1,v_2) \, g(u_1, u_2)$$
 A: 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 $$((v_1, w_1), (v_2, w_2)) \mapsto g(v_1, v_2) + h(w_1, w_2)).$$
A: If you multiply $v_1$ by $\lambda$ and $\alpha_1$ by $1/\lambda$, you don't change their tensor product $v_1 \otimes \alpha_1$. Hence any formula that one proposes for the scalar product must also be left unchanged by such an operation, and this rules out your formula with "plus" instead of "times".
A: I think it might be interesting to consider why the scalar product is defined that way: 
Recall that the tensor product $V\otimes W$ is constructed by taking $\mathrm{Free}(V\times W)$, the vector space freely generated by the elements in $V\times W$, and dividing out by the subspace $N_{V\otimes W}\subset \mathrm{Free}(V\times W)$ of all elements generated by $(v,w) + (v',w) - (v+v',w)$, and another two or three relations that you can easily find.
If you have any two maps $f\colon V\to V'$ and $g\colon W\to W'$, then it is certainly natural to consider the product map $(f,g)\colon V\times W \to V'\times W', (v,w)\mapsto (f(v),g(w))$.  This extends to a linear map on
$$ (f,g)\colon \mathrm{Free}(V\times W) \to \mathrm{Free}(V'\times W') \;, $$
because it is already defined on the generators of $\mathrm{Free}(V\times W)$.
To show that $(f,g)$ descends to a well-defined map
$$ f\otimes g\colon V\otimes W \to V'\otimes W' $$
it suffices to show that equivalence classes on the left are mapped to equivalence classes on the right, which boils down to verifying identities like
$$ (f,g) \bigl((v,w) + (v',w) - (v+v',w)\bigr) = (f,g) \bigl((v,w)\bigr) + (f,g)\bigl((v',w)\bigr) - (f,g)\bigl((v+v',w)\bigr) = (fv,gw) + (fv',gw) - (f(v+v'),gw) =  (fv,gw) + (fv',gw) - (fv+fv',gw) \;.$$
Since $(f,g)\bigl(N_{V\otimes W}\bigr) \subset N_{V'\otimes W'}$, we have a well-defined $f\otimes g$; linearity is automatic.
I haven't done the exercise, but there is no doubt that the definition of the scalar product on a tensor space follows directly from the analogue of the construction above for bilinear maps on tensor products.
