Locally in a chart, a tensor field looks like
$$T= T^{i_1,...i_n}_{j_1,...,j_m} dx^{j_1} \otimes...\otimes dx^{j_m} \otimes \partial_{i_1} \otimes ... \otimes \partial_{i_n},$$
where $T^{i_1,...i_n}_{j_1,...,j_m}$ are smooth functions.
Now, I want to see that this is then a $C^{\infty}$ multilinear map $T: \Omega_1(M) \times... \Omega_1(M) \times \Gamma(TM) \times... \times \Gamma(TM) \rightarrow C^{\infty}(M,\mathbb{R}).$
The converse direction is actually clear to me.
Now assume we would plug in a vector field multiplied by a function in the first component of the local tensor field representation (which corresponds to a tensor $dx^{j_1}$, then
there would be a term $dx^{j_1}(f \partial_i)$, since $\partial_i$ is a locally a basis of tangent spaces and $f$ is here a smooth function. Now, how is this $C^{\infty}$ linear? Or is this defined to be $f dx^{j_1}(\partial_i)$? The thing is that I don't think so, i.e. $\partial_i$ is of course not(!) $C^{\infty}$ linear (as it is a derivative), so $\partial_{i_k}(f dx^{i})\neq f \partial_{i_k}(dx^{i})$ as far as I know.
By the way: I know that $dx^i$ is just the dual basis element to $\partial_i$, but I don't get this $C^{\infty}$ linearity.
Edit: Thinking about it, I feel as if a pointwise argument could save me here:
$\partial_{i_k}(f dx^{i})|_p = \partial_{i_k}|_p(f(p)dx^{i}(p)) = f(p) \delta_{i_k,i}$ for all points $p \in M$. So it holds for the smooth function.
Is this the trick? If so, then this notation is totally confusing in this case. But maybe somebody could still explain why the vector field $\partial_{i_k}$ does in this case not act on $f$, cause I still feel as if I have not understood what I am doing here.?
I hope this answer helps you:
You are right but you are wrong, I mean, you are right because vector fields are derivations (or at least in one-to-correspondence with they), so
$$ \partial_i (fdx^j) \neq f\partial_i(dx^j). $$
However, when you define a tensor as a multilinear map, you are using (implicitly) the canonical isomorphism between a finite-dimensional vector space $V$ and its bidual $V^{**}$, so you actually are considering the $C^\infty(M)$-linear maps
$$ dx^i: TM \longrightarrow C^\infty(M) $$
and
$$ \bar{\partial}_j : T^*M \longrightarrow C^\infty(M). $$
However, since each $\bar{\partial}_j$ is in one-to-one correspondece with the vector field $\partial_j$, we usually write the vector field instead of its associated linear map. But remember, as a $C^\infty(M)$-linear map, a (1,1) tensor (for example) should be a section $T\in \Gamma^\infty(T^*M\otimes T^{**}M$).
So, summarazing, the step that you lacked was to consider the canonical isomorphism between $T_pM$ nad $T^{**}_pM$ and hence between $TM$ and $T^{**}M$.