The function $m: C^{\infty}(M) \times C^{\infty}(M) \to C^{\infty}(M)$ $m(f,g) = fg$ is $C^{\infty}$ bilinear ... so is it tensor like?

Question

The function $m: C^{\infty}(M) \times C^{\infty}(M) \to C^{\infty}(M)$ $m(f,g) = fg$ is $C^{\infty}$ linear in both components... so is it a tensor of some kind? I know (I think) it is not a multilinear map from $X(TM) \times ... X(TM) \times X(TM^*) \times... X(TM^*)$, so it isn't a tensor in the usual sense.

I would appreciate it if someone could clarify this. I am confused because I saw the Leibniz rule (say for a (1,1) tensor) for connections presented as $\nabla_X(T(\omega,Y)) = \nabla_XT(\omega,Y) + T(\nabla_X \omega,Y) + T(\omega, \nabla Y)$... and I was trying to understand how the usual Leibniz rule followed from this.

I'm confused...

For some affine connection $\nabla$, is $\nabla_{v} m = 0$ since multiplication is a "constant" operation? (Not that $\nabla_{v} m$ has any immediate meaning to me...)

Thanks!

Edit: (The Leibniz rule for covariant derivatives I am familiar with follows without from this definition after a long computation using the definition ($T \otimes S(...) = T(...)S(...)$). This is not exactly an answer to the question I asked, but it does resolve the origin of my confusion.)