Basis-free formula for $\mathrm{Hom}_k(V,V)\rightarrow V^*\otimes V$ Let $V$ be a finite dimensional vector space over a field $k$.  Then  there is a natural map $\phi:V^*\otimes V\rightarrow \mathrm{Hom}_k(V,V)$ given by $$\phi:f\otimes v\mapsto \Big(x\mapsto f(x)v\Big)$$ (extended by linearity). It's not hard to check that this is well-defined and injective, hence surjective by a dimension count.
The above suffices to define a (canonical) map $\phi^{-1}\colon\mathrm{Hom}_k(V,V)\rightarrow V^*\otimes V$.  
Explicitly, $\phi^{-1}(g)$ is the unique element of $V^*\otimes V$ that maps to $g$ under $\phi$.
This definition of $\phi$ is basis-free, but it does not seem to give a formula for $\phi^{-1}$ in the same sense that the displayed equation gives a formula for $\phi$. 
Question 1: How should I make precise the notion that the displayed equation for $\phi$ counts as a "formula", but the definition of $\phi^{-1}$ does not?
Question 2:  Is there some alternate basis-free definitionof $\phi^{-1}$ that clearly would count as a formula?
Comment:  It is easy to write down a basis-dependent formula for $\phi^{-1}$ and then prove that the result of this formula is independent of the choice of basis.  But I'm looking for a formula that never requires picking a basis in the first place.
 A: Strictly speaking, it is not true that the definition of $\mathrm{Hom}_k(V,W)\xrightarrow{\phi^{-1}} V^*\otimes W$ based on surjectivity of the injective $V^*\otimes W\xrightarrow{\phi}\mathrm{Hom}_k(V,W)$ is a basis-free definition.
This is because the surjectivity is equivalent to the assertion "$W$ is finite-dimensional", which is secretly the assertion "$W$ has a finite basis".
Here I am distinguishing between basis-free and basis-independent: the former can be defined without reference to a basis, the latter produces something out of a basis, but applied to any basis produces the same thing (although the axiom of choice obscures the distinction by giving every vector space a basis, thus effacing the wider applicability of basis-free definitions).
With that in mind, when you say that $\phi^{-1}(g)$ is the "unique element" of $V^*\otimes W$ that maps to $g\in\mathrm{Hom}_k(V,W)$ under $\phi$, what you're really doing is giving a basis-independent definition of $g$, because that's what dimension-counting does (produces elements from a basis independently of the basis you start with). Accordingly, there are always formulas associated with basis-independent definitions, it's just that the formulas have an additional parameter that is some basis, whereas basis-free definitions give formulas that depend solely on the data of the vector spaces.
So I doubt that there is a basis-free formula for $\phi^{-1}$ because the existence of $\phi^{-1}$ is equivalent to an assertion of finite-dimensionality, hence to the existence of some finite basis. Furthermore, I don't know that injectivity of $V^*\otimes W\xrightarrow{\phi}\mathrm{Hom}_k(V,W)$ has a basis-free proof, which would mean that $\phi^{-1}$ cannot even exist, let alone be defined, without a basis for $W$.
Also, the finite-dimensionality is almost a red herring. The correct thing to do is to describe the image of $V^*\otimes W\xrightarrow{\phi}\mathrm{Hom}_k(V,W)$ as consisting of those linear maps $V\to W$ whose image is finite-dimensional. Then any choice of basis $\{w_i\}$ for $W$ allows you to define $\phi^{-1}$ as follows. By definition, a $V\xrightarrow{g}W$ in the image of $V^*\otimes W\xrightarrow{\phi}\mathrm{Hom}_k(V,W)$ has finite-dimensional image, hence can be written (uniquely) as (jointly finite) sums $g(v)=\sum_i w_i^*(g(v)))w_i$ where $\{w_i^*\}\subset W$ are the dual vectors to the basis $\{w_i\}$ given by $w_i^*(w_j)=\delta_{ij}=\begin{cases}1&i=j\\0&i\neq j\end{cases}$. This says exactly that $\phi^{-1}(g)=\sum_i(w_i^*\circ g)\otimes w_i$, which is a perfectly good, unambiguous formula, albeit one that depends on the basis $\{w_i\}$.
Note that when I write $\phi^{-1}$, I mean that $\phi\circ\phi^{-1}=\mathrm{id}$. Indeed, $\phi(\sum_i(w_i^*\circ g)\otimes w_i)(v)=\sum_iw_i^*(g(v))w_i=g(v)$. We need to check that $\phi^{-1}\circ\phi=\mathrm{id}_{V^*\otimes W}$ to conclude we have an isomorphism (dimension-counting doesn't work because $V^*\otimes W$ could be infinite-dimensional). Since an arbitrary element in the image of $\phi$ is of the form $\phi(\sum_j f_j\otimes w_j)=\sum f_j\cdot w_j$ where $\{f_j\}\subset V^*$ is some set of linear functionals, we indeed have $\phi^{-1}(\sum f_j\cdot w_j)=\sum_iw_i^*\circ(\sum f_j\cdot w_j))\otimes w_i=\sum_if_i\otimes w_i$.
