Two notions of a tensor The way I've been learning tensors—building towards differential forms—has the tensor product for vectors $u,v\in V$ (say, finite-dimensional and over some field $\mathbb F$) as
$$u\otimes v=[(u,v)]\in F(V\times V)/U$$
where $F(V\times V)$ is the free vector space over $V\times V$ and $U$ the appropriate "linearizing" subspace such that the elements of $F(V\times V)/U$ are multilinear. Then $u\otimes v$ can be regarded as the map
$$u\otimes v:V^\ast\times V^\ast\to\mathbb F$$
with $(f,g)\mapsto f(u)g(v)$.
But when I'm running into tensor products in practice, I frequently see definitions with respect to some basis $\{e_1,\ldots,e_n\}$ such that for $u=u^ie_i$ and $v=v^ie_i$:
$$u\otimes v=\begin{pmatrix}u_1v_1&\cdots&u_1v_n\\
\vdots&\ddots&\vdots\\
u_nv_1&\cdots&u_nv_n\end{pmatrix}.$$
I'm told the latter case naturally injects (or maybe is isomorphic) to the former, but I'm not seeing it. Among other things, the latter $u\otimes v$ is a linear mapping $V\to V$ whereas the former $u\otimes v$ is a multilinear mapping $V^\ast\times V^\ast\to\mathbb F$.
Am I misunderstanding something fundamental, incorrectly understanding some definition, or missing some obvious correspondence?
 A: The isomorphism is not really natural, since the latter representation of $u\otimes v$ depends on the choice of basis.

These are different interpretations of tensors. The isomorphism between them is as follows:
Let $\left\{e_i\right\}_i$ be the basis for $V$. Then we have a (non-natural) isomorphism $\phi\colon V\to V^*$, where each $e_i$ is taken to the dual linear functional given by
$$e_i^*(e_j)=\begin{cases}1&\text{, if }i=j\\0&\text{, otherwise}\end{cases}$$
We can proceed with the same type of isomorphism, but now at the level of $V^*$: Take the isomorphism $\phi^*\colon V^*\to V^{**}$. So composing we obtain an isomorphism $\phi^*\circ\phi\colon V\to V^{**}$ (which is natural in the sense that it does not actually depend on the initial basis $\left\{e_i\right\}_i$, but this doesn't matter at the moment).
Now, you have also probably seen the following isomorphism: given vector spaces $E,F$, let $L(E,F)$ be the vector space of linear maps $E\to F$. Given another vector space $W$, let $Bil(E\times W,F)$ be the vector space of bilinear maps $E\times W\to F$.
Then we have the isomorphism
$$\Psi\colon Bil(E\times W,F)\to L(W,L(E,F))$$
where for each $T\in Bil(E\times W,F)$, the map $\Psi(T)\colon W\to L(E,F)$ is given by $\Psi(T)(w)(e)=T(e,w)$.
Finally, you have probably seen that if $E$ is isomorphic to another space $\tilde{E}$, then $L(E,F)$ is isomorphic to $L(\tilde{E},F)$, and similarly for $Bil(E\times W,F)$, for $F$ and $\tilde{F}$, etc. If not, this is a nice exercise.
Cool. Let's put all of this together: the former tensor is an element of $Bil(V^*\times V^*,\mathbb{F})$; The latter tensor is an element of $L(V,V)$. But we have isomorphisms
\begin{align*}
Bil(V^*\times V^*,\mathbb{F})&\cong L(V^*,L(V^*,\mathbb{F}))\\
&= L(V^*,V^{**})\\
&\cong L(V^*,V)\\
&\cong L(V,V)
\end{align*}
The latter interpretation of tensors is just the element of $L(V,V)$ which comes from the former interpretation of tensors in $Bil(V^*\times V^*,\mathbb{F})$ under the isomorphisms above.
