Laws of Exponents for Cosets - Fraleigh p. 142 14.22 
A torsion group is a group all of whose elements have finite order. A group is torsion free if the identity is 
  the only element of finite order. Prove: if G is a torsion group, then so is G/H for
  every normal subgroup H of G.
Proof: Let $xH ∈ G/H$. Because G is a torsion group,  $x^m = e$ in G for some positive integer m. Compute
  $(xH)^m$ in G/H using the representative $x$: $\color{magenta}{(xH)^m = x^mH} = eH = H$,
  so $xH$ is of finite 
  order. Because xH can be any element of G/H, we see G/H is a torsion group.

(1.) Can someone please flesh out $\color{magenta}{(xH)^m = x^mH}$? I understand $(xH)^m = (xH)...(xH)$ (m times). But the question doesn't presuppose $G,H$ are commutative, hence this is impossible: $(xH)...(xH) = \underbrace{x...x}_{m \;times}\underbrace{H...H}_{m \;times} = x^mH^m$ ?
(2.) What's the intuition?
(3.) user1729's Method 2 is  $\color{brown}{\text{definition of normality :}} g\color{brown}{H}hH=g\color{brown}{(h \quad Hh^{-1})}hH=gh \; H$.
Where does this trick of substituting H spring from? How do you predestine to do this?
(4.) user1729 says it doesn't matter that $\ker \phi = H$? Why not? Doesn't Fundamental Homomorphism Theorem require this? 
(5.) What are the sufficient conditions for $[Hg]^n = Hg^n$?
Adam Saltz's answer says G H   Abelian. But user1729's answer says G normal?
 A: Method 1: The best way, I think, to approach this problem is to forget about cosets and instead think about homomorphic images. We can do this because of the first isomorphism theorem. So, $G/H$ is just some group, $K$ say, and we have a homomorphism $\phi: G\rightarrow K$ with $\ker\phi=H$. All your question is asking is the following.

Is it true that $(\phi(g))^n=\phi(g^n)$?

And, of course, this is true because $\phi(g)\phi(h)=\phi(gh)$ for all $g, h\in G$ (as $\phi$ is a homomorphism).
Method 2: If you do still want to think about cosets then induct on $n$ using the following trick.
$$gHhH=g(hHh^{-1})hH=ghH$$
This holds because $H$ is normal in $G$.
Comments:  The OP asked me about the intuition here, and about what would happen if $H$ was abelian. The intuition is the homomorphism. The trick I talk about holds because $H$ is normal. Indeed, it holds precisely when $h$ is contained in the normaliser of $H$ (the normaliser is the largest subgroup of $G$ in which $H$ forms a normal subgroup). It is possible for this trick to not hold when $H$ is abelian - the structure of $H$ is irrelevant, rather, it is how it lies in $G$ which is important. For example, consider a dihedral group $\langle a,b;a^2,b^n,aba=b^{−1}\rangle$. Then $\langle a\rangle$ is abelian but not normal for $n>2$.
Note also that in the question the OP claims that I am saying that $G$ is normal. No I am not. I am saying that $H$ is normal in $G$. The result holds for all $H$ when $G$ is an abelian subgroup, but this is a special case because all subgroups of an abelian group are normal.
A: 
Original question (2.) Is this true on the whole: $\text{If H is any subgroup of any group G, then }[Hg]^n = Hg^n$

Let me answer your second question first.  That statement is false.  Take the free product on generators $a, b$.  Let $A$ be the subgroup generated by $a$.  Then $(Ab)^2 \neq A^2b^2$.

Original question (1.) Can someone please flesh out $T(G)g^n = (T(G)g)^n$ from?

Make sure you read the original question: the group $G$ is assumed to be abelian.  It is true that $(Hg)^n = Hg^n$ for an abelian group $G$ with subgroup $H$ and $g \in G$.  To see this, note that $(Hg)^n$ is the set of elements of form $h_1gh_2g\cdots h_k g$ with $h_i \in H$ for all $i \in \{1, k\}$.  As $G$ is abelian, we can rearrange this to $h_1h_2\cdots h_kg^n$, and $h_1h_2\cdots h_k \in H$ because $H$ is a subgroup.
A: *

*I think the easier way (thus also answering the question 2 about intuition) is to note that $x\mapsto xH$ is a group homomorphism $\phi\colon G\to G/H$. Therefore $x^m=1$ implies $\phi(x)^m=\phi(x^m)=\phi(1)=1$. At the basic level, even though $xh\ne hx$ in general, we are given that $H$ is normal, hence $xhx^{-1}\in H$ for all $h\in H$; which is equivalent ot $xH=Hx$.

*see above

*To make use of the definition of things involved should be among the first things to try. Actually, in these basic problems this method makes it often impossible to proceed in any direction other then towrds the goal.

*Well, that's debatable. In answer 1., we only used that the elements of the group $\phi$ maps to are the elements $xH$ we are interested in. We do not actually make use of the fact that $\phi(x)=1$ iff $x\in H$. But of course this is exactly how $G/H$ is constructed ...

*Note that $(Hg)^n=g^{-1}(gH)^ng$, so the situation is the same for left and right cosets. Normality is sufficient, as seen in 1. or other proofs of the original problem. (If $G$ is abelian, then all subgroups are normal, so $G$ being abelian is of course also sufficient). Normality is also necessary: Let $g\in G$. If $x\in gHg^{-1}$, then $xg^2\in (gH)^2=gHg^{-1}g^2H$, so for $(gH)^2=g^2H$ we need $xg^2\in g^2H$, i.e. $x\in g^2Hg^{-2}$. We conclude $gHg^{-1}\subseteq g^2Hg^{-2}$ for all $g\in G$. By conjugation $H\subseteq g H g^{-1}$ for all $g\in G$ and finally $H=gHg^{-1}$, i.e. $H$ is normal.
