Let $G$ be a group. Denote normal subgroup by $\triangleleft$ and subgroup by $<$.
Let $N\triangleleft G$.
Let $U\triangleleft H< G$.
Show that $NU\triangleleft NH$.
I think that I successfully worked it out below. But comparing $U\triangleleft H$ and $NU\triangleleft NH$ it really seems like we are "just multiplying by $N$ on the left of both sides", and I therefore wonder if perhaps there is some more simple way of showing it.
My attempt: First note that $NU \subset NH$. To show that $NU$ is a group the only nontrivial thing to check is that $NU$ is closed under the group operation. ($n\in N$, $h\in H$ and $u\in U$, with primes or without.)
$$\begin{align} (nu)(n'u') &= u(u^{-1}nu)n'u' \\ &= un''n'u'\\ &= un'''u' \\ &= (un'''u^{-1})uu' \\ &= n''''u'' \\ &\in NU \end{align}$$
So $NU$ is a group. In similar fashion $NH$ is shown to be a group. To show that $NU$ is normal in $NH$, we can show that for any $n\in N$ and any $h\in H$ it holds that $(nh)^{-1}NU(nh) = NU$.
$$\begin{align} (nh)^{-1}NU(nh) &= h^{-1}n^{-1}NU(nh) \\ &= h^{-1}NU(nh) \\ &= N(h^{-1}Uh)(h^{-1}nh) \\ &= NUn' \end{align}$$ Since $n'\in NU$ it follows that $NUn' = NU$. Thus $NU$ is normal in $NH$.
The proof seems correct.
As you remarked, one should first show that, whenever $A$ is a normal subgroup and $B$ is a subgroup of $G$, then $AB$ is a subgroup; indeed $$ (a_1b_1)(a_2b_2)^{-1}= \bigl( \underbrace{a_1(b_1a_2b_1^{-1})}_{\in A} \bigr) \bigl(\underbrace{b_1b_2}_{\in B}\bigr) $$ because $b_1a_2b_1^{-1}\in A$. Also $(ab)^{-1}=b^{-1}a^{-1}=(b^{-1}a^{-1}b)b^{-1}\in AB$. Similarly $BA$ is a subgroup and this has the obvious consequence that $AB=BA$.
Therefore, both $NU$ and $NH$ are subgroups of $G$. The fact that $NU\subseteq NH$ is clear.
For normality, it's easier with the inverse on the right side: $$ (n_1h_1)n_2u(n_1h_1)^{-1}= \bigl( \underbrace{n_1(h_1n_2h_1^{-1})}_{\in N} \bigr) \bigl( \underbrace{h_1uh_1^{-1}}_{\in U}\bigr)n_1^{-1} $$ and you're done because $UN=NU$.