Is this the correct subgroup of order 12 I found? Consider the following exercise:
Suppose that $H$ is a normal subgroup of $G$. IF $|H|=4$ and $gH$ has order $3$ in $G/H$, find a subgroup of order 12 in $G$.
After trying a few things that failed I decided I was going to attempt to show the following:
Claim: $K=H \sqcup gH \sqcup g^2 H$ is a subgroup of $G$ of order 12.
It's clear that $e \in K$. Also, if $x \in K$ then either $x \in H, x \in gH$ or $x \in g^2 H$. If $x\in H$ then $x^{-1} \in H$. If $x \in gH$ then $x = gh$ for some $h\in H$ and $x^{-1} = h^{-1}g^{-1} = h^{-1}g^2 = Hg^2 = gH$. Similarly for the case where $x \in g^2 H$.
Finally, I should show that $x,y \in K$ implies $xy \in K$. I can do it in a similar way I did $x^{-1}$ but I feel stupid because there are so many cases and I think there should be a neater proof than what I'm doing. 
I just don't see it.
So please:

Is there a neat(er) way of answering this exercise?

 A: You could perhaps write something like this to take care of all those cases at once:
Let $x,y\in K$, so $x=g^ih$ and $y=g^jh'$ for some $0\leq i,j\leq 2$, and $h,h'\in H$. Then
$$
xy=g^ihg^jh'\in g^iHg^jH=g^ig^jHH=g^{i+j}H\subseteq K
$$
where $Hg^j=g^jH$ since $H$ is normal, this is just the operation on the quotient group $G/H$.
A: Ben West's answer is IMO the way to go. Because the set $K$ is finite, and contains the neutral element, it suffices to show that it is closed under the group operation (this is one of the versions of the subgroup criterion).
I still want to remark on one step of your attempt that caught my eye.
There was an error in your argument at the step where you wrote $h^{-1}g^{-1}=h^{-1}g^2$. It looks like you thought that the element $g$ itself is of order three. But the only piece of information given to you was that the coset $gH$ is of order three in the quotient group. What this means is that
$$
(gH)^3=g^3H=1_{G/H}=1\cdot H=H.
$$
From this you can conclude that $g^3=h'$ for some $h'\in H$.
You need to use the normality of $H$. So if $x=gh$ this implies that
$$
x=gh=(ghg^{-1})g=h''g
$$
for some $h''\in H$. Then you can argue that
$$
x^{-1}=g^{-1}h''^{-1}=g^{-1}(g^3h'^{-1})h''^{-1}=g^2h'^{-1}h''^{-1}\in g^2H.
$$
But, you're right. This leads to you needing to check many cases. Back to Ben's answer.
A: With slightly more sophistication: apparently $3\mid[G:H]$, hence $3 \mid|G|$. By Cauchy's Theorem, there must be an element $x \in G$ of order $3$. Put $K=\langle x \rangle$. Since $H$ is normal, $HK$ is a subgroup of $G$ and $|HK|=\frac{|H||K|}{|H \cap K|}=\frac{4 \cdot 3}{1}=12$.
A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Perhaps one could add that the correspondence theorem yields that the pre-image $K  = \pi^{-1}(\Span{g H}) \ge H$ of the subgroup $\Span{g H}$ of order $3$ of $G/H$ under the canonical homomorphism
$$
\pi : G \to G/H
$$
is indeed a subgroup of order $12$ of $G$.
This is because, by the above theorem,

if $L \le \dfrac{G}{H}$, then $L = \dfrac{\pi^{-1}(L)}{H}$, where $\pi^{-1}(L) = \Set{x \in G : \pi(x) \in L }$ is a subgroup of $G$ containing $H$. 

In this particular case we have thus
$$
\Span{g H}
=
\frac{\pi^{-1}(\Span{g H})}{H}
=
\frac{K}{H},
$$
and thus using Lagrange's theorem,
$$
3 =
\Size{\Span{g H}} = \Size{\frac{K}{H}} = \frac{\Size{K}}{\Size{H}}
= \frac{\Size{K}}{4},
$$
so that $\Size{K} = 12$.
By the way, as noted by OP, $K = \Span{g, H}$ here.
