I don't have your notes in front of me but I'm pretty sure that applies only when you have multiple observations. I mean think about it. Someone tells you X=x and asks you to test for $H_0$ vs. $H_a$. What are you going to sum?

I don't know how you are being taught this stuff, I'm not in any sort of communication w/your teacher. The way I would approach this problem is as follows. This is upside down from the usual NP but is easier in this case and entirely equivalent.

find $T=L(H_a)/L(H_0)$

In the first bit of your problem this is

$T=\dfrac{2 x^{2-1}}{1 x^{1-1}}= 2x$

The factor of 2 doesn't add any value here so just let

$T=x$

Now by NP your test is that $T\geq \eta$ to declare $H_a$

The Type I error will be given by

$Pr[\text{select } H_a|H_0]=\alpha$

In this case this is

$Pr[x\geq \eta | H_0]$

if you work it out this is just

$1-\eta, ~~0\leq \eta \leq 1$

so letting $1-\eta =\alpha \Rightarrow \eta = 1-\alpha$ we can then solve for the power of the test in terms of the Type I error

$Pr[x\geq (1-\alpha) |H_a]=$

$\displaystyle{\int_{(1-\alpha)}^1} 2x ~dx=1-(1-\alpha)^2$ for $0 \leq (1-\alpha) \leq 1$

and from this you can plot the ROC curve for for this test, i.e. the Power vs. the Type I error