If $V$ is finite-dimensional, choose a basis $\{ |b_{1} \rangle, \dots, |b_{n} \rangle \}$. There's a dual basis $\{ f_{1} , \dots, f_{n} \}$ for $\widetilde{V}$, which is the same as $\mathcal{T}^{1}(V)$. So, $f_{j_{1}} \otimes \dots \otimes f_{j_{r}} \in \mathcal{F}^{r}(V)$ where $j_{1}, \dots , j_{r}$ are integers selected from $\{ 1, \dots, n \}$. Now Any $T \in \mathcal{F}^{r}(V)$ can be written as $T = \sum\limits_{j_{1}, \dots, j_{r} =1}^{n} T\big(|b_{j_{1}}\rangle , \dots, |b_{j_{r}}\rangle \big) \left( f_{j_{1}} \otimes \dots \otimes f_{j_{r}}\right)$. More succinctly, $\text{Span}\big( \{ f_{j_{1}} \otimes \dots \otimes f_{j_{r}}\} \big) = \mathcal{F}^{r}(V)$. To establish linear independence consider a linear combination of the form $\sum\limits_{j_{1}, \dots, j_{r} =1}^{n} c_{j_{1}, \dots, j_{r}} \left( f_{j_{1}} \otimes \dots \otimes f_{j_{r}}\right) = 0$. Given the way the dual basis is defined, $(f_{j_{1}} \otimes \dots \otimes f_{j_{r}})\big(|b_{j_{1}} \rangle, \dots, |b_{j_{r}} \rangle\big) = 1$, but it is $0$ for any other possible list of basis vector inputs. So if we plug $|b_{j_{1}} \rangle, \dots, |b_{j_{r}} \rangle$ into the linear combination above, the result is $c_{j_{1}, \dots, j_{r}} = 0$.
I am reading this proof and I can follow it up to the point where they state that $$\sum\limits_{j_{1}, \dots, j_{r} =1}^{n} c_{j_{1}, \dots, j_{r}} \left( f_{j_{1}} \otimes \dots \otimes f_{j_{r}}\right) = 0.$$
I'm lost at what this $c_{j_{1}, \dots, j_{r}}$ is supposed to represent. Is it that $$c_{j_{1}, \dots, j_{r}}= c_{j_1} \cdot c_{j_2} \cdots c_{j_k}$$ or is it just one scalar?
The coefficient $c_{j_1,\dots, j_r}$ is a scalar (more precisely, an element of the base field). That is, for each choice of $j_1,\dots,j_r\in\{1,\dots,n\}$, there is a single scalar $c_{j_1,\dots,j_r}$. It does not denote a product of scalars.