Answer 1. The closest thing to this construction I have seen is the Eilenberg-Watts theorem, which says that for any right exact functor $F\colon R$-mod$\to Ab$ that commutes with arbitrary direct sums, we have a natural isomorphism $F(-)\cong F(R)\otimes_R-$, where $F(R)$ is given its natural structure as a right $R$-module.
The key observation to Eilenberg's original proof is that given an $R$-module $M$, the canonical module homomorphism $\bigoplus_{m\in |M|}R\twoheadrightarrow M$ is in fact an $R$-bilinear function when considered as a two-variable function, and that consequently so is the image $\bigoplus_{m\in|M|}F(R)\twoheadrightarrow F(M)$ of the map under $F$. Then a little bit of diagram chasing shows that the induced map $F(R)\otimes_R M\to F(M)$ is in fact an isomorphism.
Hence, you can obtain constructions of the tensor product $M\otimes_R N$ from any right exact, direct-sum preserving functor $F_M$ for which $F_M(R)=M$. Thus, one should not be surprised at there being a ton of different constructions of the tensor product.
It is not the result of the theorem that's relevant here, however, but rather the idea behind the proof. Adapting, it seems to boil down to the observation that $\bigoplus_{m\in|M|} N$ has $|M|_{Ab}\otimes_\mathbb Z N$ as the natural quotient by the additive relations of $|M|_{Ab}$ that $|M|_{Set}$ has forgotten, and that $\bigoplus_{n\in|N|} M$ has $M\otimes_\mathbb Z |N|_{Ab}$ as the natural quotient by the additive realtions of $|N|_{Ab}$ forgotten by $|N|_{Set}$ (the proofs of these facts should be the same as in Eilenberg's proof). Then all your construction does is realize $M\otimes_R N$ as the pushout of the two.
Answer 2. I do not know of a textbook that does this stuff.
Answer 3. What is wrong with free modules? Your distaste for them mystifies me since I perceive algebraic objects are by any reasonable definition algebraic by virtue of being given as quotients of free objects (that's what an equation is). If I were teaching, what I would do is show how the (classical) explicit construction of the tensor product is nothing more than expanding the definitional hom-tensor adjunction, the internal hom to hom-set relationship, and the cartesian product to set-hom adjunction. Since I do not think the explicit construction is ever helpful for computational purposes, I would focus on the categorical properties from which one can both deduce the construction, and actually use for computation (e.g. preservation of direct sums and right exactness).
We have:
$$2 [(1 - \omega) \otimes (1 + \omega)] = 2(1 - \omega) \otimes (1 + \omega) = 2 \otimes (1 - \omega) (1 + \omega) = 2 \otimes 6 = 2 [2 \otimes 3].$$
Similarly,
$$3 [(1 - \omega) \otimes (1 + \omega)] = (1 - \omega) \otimes 3 (1 + \omega) = (1 - \omega) (1 + \omega) \otimes 3 = 6 \otimes 3 = 3 [2 \otimes 3].$$
Therefore, subtracting the two gives:
$$(1 - \omega) \otimes (1 + \omega) = 2 \otimes 3.$$
A possible brute force method to solve the problem would be:
First, $I$ is generated as an Abelian group by $1-\omega, \omega(1-\omega) = 5 + \omega, 2, 2\omega$. Now, you could find the relations between these elements by finding the kernel of the matrix $A = \begin{bmatrix} 1 & 5 & 2 & 0 \\ -1 & 1 & 0 & 2 \end{bmatrix}$ (as a subgroup of $\mathbb{Z}^4$). One way to do this would be to use a Smith normal form calculation to write $A = P \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \end{bmatrix} Q$ for invertible matrices $P, Q$, and then the kernel of $A$ would be generated by $Q^{-1} e_3, Q^{-1} e_4$.
If I'm not mistaken (though I very well could be), this should give presentations $I \simeq \langle a, b \mid 2a = (1 - \omega) b, (1 + \omega) a = 3 b \rangle$ and $J \simeq \langle c, d \mid 3c = (1 + \omega) d, (1 - \omega) c = 2d \rangle$ for $I$ and $J$ as $R$-modules. Therefore, $I \otimes_R J$ would have a presentation in terms of 4 generators $a \otimes c, a\otimes d, b\otimes c, b\otimes d$, and eight relations $2a\otimes c = (1-\omega)b \otimes c, \ldots, (1-\omega)b\otimes c = 2b\otimes d$. You now want to know whether $a\otimes c - b\otimes d$ is zero in this tensor product, which is equivalent to asking whether $a\otimes c - b\otimes d$ is in the submodule of $R^4$ generated by these relations. That question should be straightforward if tedious to answer by a Smith normal form calculation on an $8 \times 16$ matrix (using that the tensor product is generated as an Abelian group by $a\otimes c, \omega a\otimes c, \ldots$ and similarly the relations are given by the original relations along with the relations times $\omega$).
Best Answer
I wouldn't say that it follows that the map $M\otimes \bigcup_s J_s\to M\otimes R$ exists - this is just immediate: if $\sum_k m_k\otimes i_k=0$ for some $i_k\in I$, then it's easy to see that $\sum_k m_k\otimes i_k=0$ in $M\otimes R$ as well (this is true basically whenever $M\otimes I$ is well-defined).
Injectivity of $\phi_s$ does indeed follow from the injectivity of $\phi_s$: briefly, for any $\sum_k m_k\otimes i_k\neq \sum_j m'_j\otimes i'_j\in M\otimes I$, there is some $J_s$ such that all $i_k,i'_j\in J_s$, and the conclusion easily follows by considering $\phi_s$.