Can a bilinear map on smooth functions induce a bilinear map on tensor fields in this way? Given a manifold $M$ with a bilinear bracket $(\cdot,\cdot) : C^\infty(M) \times C^\infty(M) \to C^\infty(M)$ can it induce a bilinear map for pairs of tensor fields of different ranks in a way described below?
For two smooth functions $f$ and $g$ we have a third $(f,g)$. If we take, say a $1$-form $u =  u_1 \, dx_1 + u_2 \, dx_2$, on $\mathbb R^2$ (2-D for simplicity) we may define:
$$(u,f) = (u_1, f)\, dx_1 + (u_2,f)\,dx_2.$$
Similarly, for two $1$-forms $u$ and $v$:
$$(u,v) = (u_1,v_1) \, dx_1 \otimes dx_1 + (u_1,v_2) \, dx_1 \otimes dx_2 + (u_2,v_1) \, dx_2 \otimes dx_1 + (u_2,v_2) \, dx_2 \otimes dx_2.$$
Are this definitions coordinate-independent? If so, how can they be extended to arbitrary manifolds in a coordinate-free manner?
May an additional properties for $(\cdot,\cdot)$ help, for example Leibniz rule $(fg,h) = (f,h) g + f (g,h)$ and that $(\cdot,\cdot)$ is symmetric positive definite?
 A: Depends - do you mean bilinear over $\mathbb C^\infty (M)$ or over $\mathbb R$? 
If over $C^\infty (M)$, then the only such brackets are $(f,g) = hfg$ for some $h \in C^\infty (M)$, since $$(C^\infty(M) \otimes C^\infty(M))^* \otimes C^\infty(M) \simeq C^\infty(M)$$ over $C^\infty(M)$. In this case your extension works, but it is simply $$(u,v) = h u\otimes v.$$
If you mean over $\mathbb R$, then your extension is coordinate-dependent: working with two coordinate systems $x^i$ and $y^\alpha$, the expression in the $y^\alpha$ coordinates is
$$\sum_\alpha (u(\partial_\alpha), f) dy^\alpha = \sum_{\alpha,i,j} \left(\frac{\partial x^i}{\partial y^\alpha}u(\partial_i),f\right)\frac{\partial y^\alpha}{\partial x^j}dx^j,$$ which is only equal to $\sum_i (u(\partial_i),f)dx^i$ when we can pull the $\partial x^i /\partial y^\alpha$ out of the bracket. Thus if you want your extension to be coordinate-independent for all tensors then it must be at least bilinear over the functions that can arise from coordinate changes, which is an awful lot of functions - so many that I'm pretty sure it forces you into $C^\infty (M)$-bilinearity.
