Showing that if a subgroup is normal, it's surjective homomorphic image is normal

To show that $\alpha(U)$ is a normal subgroup, you need to prove that $\alpha(U) = x \alpha(U) x^{-1}$ for all $x \in H$. But any $x \in H$ can be written in the form $\alpha(g)$ for $g \in G$ since $\alpha$ is surjective. Thus, you need only prove that $\alpha(U) = \alpha(g) \alpha(U) \alpha(g)^{-1}$ for all $g \in G$. Note the $-1$ exponent is "on the outside", so you really do need to take that last step as you suspected.

And yes, as written it is meant that you should prove that $\alpha(U)$ is a subgroup. The proof requires you to carefully work through the definition/criterion for being a subgroup, but nothing beyond that.

Also, +1 for showing your reasoning and clearly indicating what you are unsure of.