If you dont have any further knowledge on $M \otimes N$, this is the very first step and an instant consequence of the universal property of the tensor-product:
$m \otimes n \in M \otimes N$ is non-zero iff there exists some $R$-Module $T$ and a bilinear map $M \times N \to T$, which maps $(m,n)$ to non-zero.
Proof:
If $m \otimes n \neq 0$, let $T = M \otimes N$ and consider the map $M \times N \to M \otimes N$ from the universal property.
On the other hand, if such a $T$ and a map exists, by the universal property, we get a map $M \otimes N \to T$, which maps $m \otimes n$ to non-zero. In particular $m \otimes n$ is non-zero.
In our case, choose $T=\mathbb Z/2\mathbb Z$. The map
$2\mathbb Z \times \mathbb Z/2\mathbb Z, (a,b) \mapsto \frac{ab}{2}$ is bilinear and maps $(2,1)$ to $1 \neq 0$, hence $0 \neq 2 \otimes 1 \in 2\mathbb Z \otimes \mathbb Z/2\mathbb Z$.
Here is a fairly simple argument directly from the universal property. Given an abelian group $A$, define a group $F$ of "fractions" $\frac{a}{n}$ where $a\in A$ and $n\in\mathbb{Z}\setminus\{0\}$. We impose an equivalence relation on these fractions by saying $\frac{a}{n}=\frac{b}{m}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(ma-nb)=0$. Using the usual formula for addition of fractions, you can check that the set $F$ of equivalence classes of fractions forms an abelian group.
There is now a bilinear map $f:A\times\mathbb{Q}\to F$ defined by $f(a,m/n)=\frac{am}{n}$. This induces a homomorphism $g:A\otimes\mathbb{Q}\to F$ which sends $a\otimes 1$ to the fraction $\frac{a}{1}$. So to show that $a\otimes 1\neq 0$, it suffices to show the fraction $\frac{a}{1}$ is nonzero. The zero element of $F$ is the fraction $\frac{0}{1}$, and $\frac{a}{1}=\frac{0}{1}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(1\cdot a-1\cdot 0)=0$, or in other words such that $ka=0$. So if $a$ is a non-torsion element, then $a\otimes 1\neq 0$. (In fact, this map $g$ is actually an isomorphism, but that takes more work to prove.)
The case of $\mathbb{R}$ follows from the case of $\mathbb{Q}$ since $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ and tensor products distribute over direct sums.
From a broader perspective, what's really going on here is that $\mathbb{Q}$ (or $\mathbb{R}$) is a flat $\mathbb{Z}$-module, meaning that if $A$ is a $\mathbb{Z}$-module and $B\subseteq A$ is a submodule, then the map $B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$ induced by the bilinear map sending $(b,q)\in\mathbb{Q}$ to $b\otimes q\in A\otimes\mathbb{Q}$ is injective. So, in particular, if $a\in A$ is any element such that $na\neq 0$ for all $n$, the submodule $B\subseteq A$ generated by $a$ is isomorphic to $\mathbb{Z}$. Since $\mathbb{Z}\otimes\mathbb{Q}\cong\mathbb{Q}$ and we have an injective map $\mathbb{Z}\otimes\mathbb{Q}\cong B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$, this implies $A\otimes\mathbb{Q}$ is nontrivial (and in fact $a\otimes1\neq 0$, since it is the image of the nonzero element $1\otimes 1\in \mathbb{Z}\otimes\mathbb{Q}$).
The argument using fractions above can be generalized to show that if $R$ is any commutative ring, then any localization of $R$ is flat as an $R$-module. For $\mathbb{Z}$-modules, flatness can be characterized quite simply in general: a $\mathbb{Z}$-module is flat iff it is torsion-free. See Show that a Z-module A is flat if and only if it is torsion free? for some sketches of the proof.
Best Answer
For your first question, $1$ does not lie in $m$, so $1 \otimes xy$ is not actually an element of $m \otimes m$.
$R$-linearity implies $a(b\otimes c)=(ab)\otimes c$, but (and this is important), if you have $(ab)\otimes c$, and $b$ does not lie in $m$, then "$ab$" is not actually a factorization of the element inside $m$, and if we try to factor out $a$, we get $a(b\otimes c)$, which is non-sensical since $b$ does not lie in $m$.
For your second question, $(x+y)x\otimes y=x((x+y)\otimes y)=(x+y)\otimes(xy)$. The reason we were allowed to take out the $x$ in this case was because $x+y$ lies in $m$, so all of the elements in the equations remained in $m \otimes m$.
Cheers,
Rofler