Not entirely sure what you're looking for so I'll just throw out some thoughts:
Recall that $V^{*}$ is the set of linear maps from $V$ to $\mathbb{R}$. Tensoring with $W$ effectively replaces the $\mathbb{R}$ with $W$. So the star on $V$ represents that the maps are coming from $V$; the lack of one on $W$ represents that you're getting elements of $W$. Likewise, I'd expect that elements of $\text{Hom}(V,W)$ would covary with a change of basis of $V$ and contravary with one of $W$ (if you're unfamiliar with these terms, here's the Wikipedia article).
For a bit more rigor, recall that you have a natural pairing between $V$ and $V^{*}$, which is just a map $$\langle -,- \rangle: V\times V^*\rightarrow\mathbb{R}$$ defined by $\langle v,\alpha\rangle=\alpha(v)$. As you can check, this is invariant under change of basis (hence the "natural") -- this actually follows from $V$ and $V$ varying oppositely. This also has the property that if $v_i,\alpha^i$ are dual bases for $V$ and $V^{*}$, then $\langle v_i,\alpha^i\rangle=1$, and $\langle v_i,\alpha^j\rangle=0$ when $i\ne j$.
So now choose bases $v_i,w_j$ for $V$ and $W$ and a dual basis $\alpha^i$ for $V^{*}$. Given an element $v=\sum b^iv_i$ of $V$ and an element $f=\sum a_i^jv^i\otimes w_j$ of $V^*\otimes W$, we get
$$f(v)=\sum_{i,j,k}(a_i^j\alpha^i\otimes w_j)(b^kv_k)=\sum_{i,j,k}(a_i^jb^k\langle v_k,\alpha^i\rangle\otimes w_j).$$
Since the pairing is only nonzero for $k=i$, this reduces to
$$f(v)=\sum_{i,j}a_i^jb^i\langle v_i,\alpha^i\rangle\otimes w_j=\sum_{i,j}a_i^jb^iw_j$$
where the tensor product has disappeared because the output of the pairing is just a real number.
But this is just a vector in $W$! And conversely, given a linear map $V\rightarrow W$, we can pick bases and write the map as a matrix $(a_i^j)$, which then transforms to the above tensor product.
I found these notes very helpful when I was learning this stuff. You can change the 5 in the URL to other numbers to read all the notes, though I believe they stop at 9.