Any linear map $C$ between two vector spaces ${\cal U}$ and ${\cal W}$ is uniquely defined as long as it is defined on a basis of ${\cal U}\,.$ A basis of ${\cal T}^r_s(V)$ is
$$\tag{1}
h_{i_1}\otimes\,...\,\otimes\,h_{i_r}\otimes \theta^{j_1}\,\otimes\,...\otimes\,\theta^{j_s}
$$
where all $i_k$ and all $j_l$ run through $\{1,...,n\}$ and
$\{h_1,...,h_n\}$ is a basis of $V$ and
$\{\theta_1,...\theta_n\}$ is a basis of $V^*\,.$ Unfortunately I do not have access to Newman's book but I strongly believe that he defines the map $C^k_l$ on a larger set than a basis of ${\cal T}^r_s(V)$ by allowing each of the factors $v_i$ and $\eta^j$ to be a linear combination of the basis vectors $h_\mu\,,$ resp. $\theta_\nu\,.$ (If his $v_i$ were all different, or his $\eta^j\,,$ this would not define $C^k_l$ on a basis of ${\cal T}^r_s(V)\,.$ Clearly, in (1) the factors are not all different.)
The only thing (if anything) that needs to be checked is if the contraction form of the right hand side of
\begin{align}
&C^k_l(v_1\otimes\ldots\otimes v_r\otimes\eta^1\otimes\ldots\otimes\eta^s)\\&=v_k(\eta^l)\,v_1\otimes\ldots\otimes \hat v_k\otimes\ldots\otimes v_r\otimes\eta^1\otimes\ldots\otimes\hat\eta^l\otimes\ldots\otimes\eta^s\tag{2}
\end{align}
is compatible with the same form for basis vectors:
\begin{align}
&C^k_l(h_{i_1}\otimes\,...\,\otimes\,h_{i_r}\otimes \theta^{j_1}\,\otimes\,...\otimes\,\theta^{j_s})\\&=h_{i_k}(\theta^{j_l})\,h_{i_1}\otimes\ldots\otimes \hat h_{i_k}\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes\hat\theta^{j_l}\otimes\ldots\otimes\theta^{j_s}\,.\tag{3}
\end{align}
By linearity it is enough to expand only the hatted factors
$$
v_k=\alpha{^\mu}_k\,h_\mu\,,\quad\eta^l={\beta^l}_\nu\,\theta^\nu\quad\text{ (using summation convention) }
$$
and assume that in (2) the un-hatted factors are basis vectors. In other words we need to check that (3) implies
\begin{align}
&C^k_l(h_{i_1}\otimes\ldots\otimes v_k\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes \eta^l\otimes\ldots\otimes\theta^{j_s})\\&=v_k(\eta^l)\,h_{i_1}\otimes\ldots\otimes \hat{v}_k\otimes\ldots\otimes h_{i_r}\otimes\theta^{j_1}\otimes\ldots\otimes \hat{\eta}^l\otimes\ldots\otimes\theta^{j_s}\,.\tag{2'}
\end{align}
From
$$
v_k(\eta^l)=\alpha{^\mu}_k{\beta^l}_\nu\,h_\mu(\theta^\nu)
$$
this should however by obvious. If not please note that $\mu,\nu$ are just dummy indices which we can replace:
$$
v_k(\eta^l)=\alpha{^{i_k}}_k{\beta^l}_{j_l}\,h_{i_k}(\theta^{j_l})\,.
$$
$\otimes$, also called tensing, is something you get bundled with the tensor product that you don't have in an ordinary vector space. How the tensor product vector space and tensing work together are what the real "meat" behind the tensor product is. Constructions are not "the real meaning", because there are an infinite number of them that will do the job - they're really better understood as first, proofs that the tensor product exists, and second, encodings of the tensor product in the medium of sets, similar to how that, on a computer, ASCII is an encoding of text in binary numbers. The same applies to constructions of most other mathematical objects using sets.
Hence, what $v \otimes w$ "is" will depend on which construction you choose. In the first case, it is not circular: we define $v \otimes w$ to be the cell in $Q$ containing the ordered pair $(v, w)$. And in general cases, that is the
The "real meaning" behind the tensor product, and that nifty little tensing operation in comes with, is that it provides a space which lets you work with bilinear maps (generically, $n$-linear maps) as though they were unilinear maps. Now, I suppose you (or some others) might be thinking, "but isn't $V \times W$ a vector space? So isn't a bilinear map $f: V \times W \rightarrow Z$, a linear map from an ordered pair $(v, w)$, viewed as a single vector in $V \times W$?" Yes, it is, but remember that a bilinear map must be linear in each argument individually, and this gives them more structure that is not captured by a simple linear map out of $V \times W$.
Hence the tensor product. We can think of this as enriching the domain so that, in this new domain, which we call $V \otimes W$, being unilinear now carries all the structural weight of being bilinear on the $V \times W$ domain.
In particular, the tensor product as the property that every bilinear map $f: V \times W \rightarrow Z$, can be understood uniquely as a unilinear map $f_\otimes : V \otimes W \rightarrow Z$, where
$$f_\otimes(v \otimes w) := f(v, w).$$
Moreover, every vector space that has this property is isomorphic to the tensor product. The construction, then, simply shows that this is not a vacuous statement, i.e. that we are actually talking about a real mathematical object here. In this regard, it's kind of like the various constructions of the real numbers: the real numbers are "really" the single object known as "the Dedekind-complete ordered field" - what those constructions do is they prove that such a thing actually exists.
In this setting, the meaning of $v \otimes w$ is that it's a "package" that wraps together $v$ and $w$ into a single matrovector for processing into a linear map in such a fashion that said linear maps acquire all the extra structure bilinear maps have, which simply taking an ordered pair would not be able to do.
Best Answer
Note that for $F \in L(V_1, \dots, V_k; W)$, we have $F_{i_1,\ldots,i_k} = F(\mathbf{e}_{i_1}^{(1)},\ldots,\mathbf{e}_{i_k}^{(k)}) \in W$. Let $\mathbf{w}_1, \dots, \mathbf{w}_m$ be a basis for $W$, then $F_{i_1,\dots, i_k} = F_{i_1,\dots, i_k}^l\mathbf{w}_l$ where $F_{i_1,\dots, i_k}^l \in \mathbb{R}$ for $l = 1, \dots, m$. So now
$$F = F_{i_1,\ldots,i_k}^l\mathbf{e}_{(1)}^{i_1}\otimes\cdots\otimes\mathbf{e}_{(k)}^{i_k}\otimes\mathbf{w}_l.$$