I think there is a bit of confusion around. The last paragraph (about the tensor characterizing lemma) does not really make sense as it stands. For a map $T_pM\times\dots \times T^*_pM\to\mathbb R$ neither being linear over smooth functions nor being tensorial makes sense. For both concepts you need an operator mapping sections to smooth functions.
Indeed I would call the equivalence you describe in the second paragraph "tensor characterizing lemma", since for an operator $SM\times\dots\times S^*M\to C^\infty M$ both being tensorial and being linear over $C^\infty M$ really makes sense. If $F$ is such an operator, then for each point $p\in M$, you can look at $F(\xi_1,\dots,\phi_\ell)(p)$ (so you evaluate the function obtained from the vector fields and one-forms in the point $p$). Now you call $F$ "tensorial" if the result is zero whenver a single $\xi$ or a single $\phi$ vanishes in the point $p$. Equivalently, this means that $F(\xi_1,\dots,\phi_\ell)(p)$ only depends on the values of the $\xi$'s and the $\phi$'s in $p$. If this is the case, then $F$ actually defines a map $T_pM\times\dots \times T^*_pM\to\mathbb R$ in each point $p$, and it is easy to see that these maps define a tensor field in the sense of your first definition.
It follows easily from the definitions the a tensorial operator $F$ is linear over $C^\infty M$ (since $(f\xi)(p)=f(p)\xi(p)$ and similarly for the $\phi$'s). The interesting part of the proof is the converse direction, which is most easily done in two steps. Assuming that $F$ is linear over smooth functions in each argument, you first prove that it defines a local operator. This means that if $U\subset M$ is open and one of the $\xi$'s or $\phi$'s vanishes identically on $U$, then the function $F(\xi_1,\dots,\phi_\ell)$ vanishes identically on $U$. To see this, suppose that $\xi_1$ vanishes on $U$ and take $p\in U$. Let $f$ be a bump function with support contained in $U$ such that $f(p)=1$. Then $f\xi_1$ vanishes identically, so $F(f\xi_1,\dots,\phi_\ell)=fF(\xi_1,\dots,\phi_\ell)$ vanishes identically since $f$ is linear. But evaluating this in $p$, you get $0=f(p)F(\xi_1,\dots,\phi_\ell)(p)=1\cdot F(\xi_1,\dots,\phi_\ell)(p)$.
Knowing this, you conclude that for any open subset $U\in M$, the restriction of $F(\xi_1,\dots,\phi_\ell)$ to $U$ depends only on the restrictions of the $\xi$'s and $\phi$'s to $U$. But given a point $p$, you can restrict to the domain $U$ of a chart containing $p$. On $U$, you can expand the $\xi$'s in terms of coordinate vector fields and the $\phi$'s in terms of coordinate one-forms. Of course, $\xi(p)=0$ if all the coefficients in such an expansion vanish in $p$. But then you just insert into $F$, pull out all the functions and conclude that $F(\xi_1,\dots,\phi_\ell)(p)=0$.
