What is the left adjoint to forgetful functor from Mod(R) to Ab I'm actually trying to fully understand an old reply here Forgetful functor from R-modules to abelian groups? but I figure my doubts can also be formulated as a separate question.
So consider the forgetful functor from the category of modules over $R$ into the category of abelian groups. I would like to understand what is the left adjoint to this functor, but I have trouble imagining what could it be. My approach to understanding is that if $G$ is the forgetful functor, $F$ is the adjoint we are looking for, and $c$ is an abelian group whereas $d$ is a module, then we want to pick $F$ in a way that there will be a natural bijection between group homomorphisms from $Hom(c,G(d)$ and module homomorphisms in $Hom(d, F(c))$. 
In the post Forgetful functor from R-modules to abelian groups? the person replying states that the left adjoint sends a group $A$ to $R \otimes_{\mathbb{Z}} A$. I'm not very familiar with tensor products and tried reading about them but this just made me more confused. So I guess my question comes down to what kind of object is $R \otimes_{\mathbb{Z}} A$ and why is it the left adjoint here? Does the answer change if $R$ is a commutative ring?
 A: The left adjoint to a forgetful functor is always some kind of free construction functor.
In this case we take the elements of the given Abelian group $A$ and make a formal possibility to multiply them (from left) by elements of our ring $R$, and add them. We can call it the $R$-module freely generated on an Abelian group.
Its elements are [formal sums of] pairs $\langle r,a\rangle$, so that we can multiply these by elements of $R$ from the left. 
Pairs of the form are considered equal: $\langle rn,a\rangle = \langle r,na\rangle=n\cdot\langle r,a\rangle$ for any $n\in\Bbb Z$. 
There are also similar constrainments about the formal sums:
$$\langle r_1+r_2,a\rangle = \langle r_1,a\rangle + \langle r_2,a\rangle \quad\quad\quad \langle r,a_1+a_2\rangle=\langle r,a_1\rangle+\langle r,a_2\rangle$$
The usual notation for such a pair $\langle r,a\rangle$ in this context is $r\otimes a$. 
This construction can be made exact using the free Abelian group generated by $R\times A$ (so that we have the formal sums of our pairs) and then define a normal subgroup based on the above relations, and quotient it out.

Note that if we skip addition from the above, this situation might be easier to comprehend first: 
Then we would have $A$ as a pure set, and $R$ as a monoid (a set with associative multiplication with unity). 
The nonadditive $R$-modules are called $R$-sets or $R$-actions, and the left adjoint of the forgetful functor from $R$-actions to sets is simply $A\mapsto R\times A$ with the $R$-action $r\cdot\langle r_1,a\rangle:=\langle rr_1,a\rangle$.
If we then have a function $f:A\to M$ with $M$ (the underlying set of) an $R$-action, it induces a morphism of $R$-actions $\varphi:R\times A\to M$:
$$\langle r,a\rangle \,\mapsto\, r\cdot f(a)$$
and conversely, such an $f$ can be recovered by taking $a\mapsto \varphi(\langle 1,a\rangle)$. 
