It seems to me, through my mathematical immaturity, that the tensor product seems to beg for more well-definition. I am working in vector spaces (so we always have a free module) and here is what my professor has shown me thus far.
We can define the tensor product of two maps (multi-linear) as follows. Let $S \in \mathcal{L}(V_1, \dots, V_n; \mathcal{L}(W;,Z))$ and $T \in \mathcal{L}(V_{n+1}, \dots , V_{n+m};W)$, We define $S \otimes T \in \mathcal{L}(V_1, \dots , V_{n+m};Z)$ by setting
$$S \otimes T(v_1, \dots ,v_{n+m})=S(v_1, \dots, v_n)[T(v_{n+1}, \dots , v_{n+m})]$$
Now, we do have $\mathcal{L}(V_1, \dots , V_{n+m};Z) \cong V^*_1 \otimes \dots \otimes V^*_{n+m} \otimes Z$ I believe. So it is, up to isomorphism, a tensor but not, itself, a tensor.
Further, suppose that $V_1, \dots , V_n$ are vector spaces. We define the tensor product
$$V_1 \otimes \dots \otimes V_n = \mathcal{L}(V^*_1, \dots V^*_n; \mathbb{F})$$
Since we regard $V$ and $V^{**}$ to be identified we have
$$v_1 \otimes \dots \otimes v_n \in V_1 \otimes \dots \otimes V_n$$
defined
$$(v_1 \otimes \dots \otimes v_n)(L_1, \dots L_n)=L_1(v_1)\dots L_n(v_n)$$
Finally, we have defined a tensor of type $m,n$ to be a multi-linear map from $\underbrace{V^* \times \dots \times V^*}_{m \text{ times}}\times \underbrace{V \times \dots \times V}_{n \text{ times}} \to \mathbb{F}$.
problem
So it seems to me that tensor products do not always produce tensors? That a tensor product sometimes is and sometimes is not a map to the field? Which makes me wonder how we can consider the idea to be well-defined? I have to be told by some to think about it in terms of the universal property, i.e., it takes multi-linear maps to linear ones but that isn't as illuminating as some may think. How is one to think about this product and these objects? Thanks for your help!