Show that if $n$ is a power of $2$, say $n = 2^k$, then we have the equality $\sum_{i=0}^{k} \lg \frac{n}{2^i} = \Theta(\lg^2 n)$.
The first step is to prove $O(\lg^2n)$: $$ \lg \frac{2^k}{2^0} + \lg \frac{2^k}{2^1} + \dots + \lg \frac{2^k}{2^k} \leq \lg^2 n. $$
My trouble is trying to figure out what sequence of the statement can relate to $\lg^2 n$. Also, I am not sure whether $\lg^2 n$ represents $\lg(\lg(n))$ or $(\lg n)^2$.
Here, $\lg^2 n$ stands for $(\log_2 n)^2$.
Your question is then to study the asymptotic behavior of $\sum_{k=0}^{\log_2 n} \log_2 \frac{n}{2^k}$.${}^{(\dagger)}$
Below are two different methods, one using knowledge of $\sum_{\ell=1}^m \ell = \frac{m(m+1)}{2}$ and yielding a sharp estimate; the second requiring no prior knowledge, but only giving a looser, $\Theta(\cdot)$-type asymptotic estimate.
Then,
$$ \sum_{i=0}^{\log_2 n} \log_2 \frac{n}{2^i} = \sum_{j=0}^{\log_2 n} \log_2 2^j = \sum_{j=0}^{\log_2 n} j = \frac{\log_2 n(\log_2 n + 1)}{2} \operatorname*{\sim}_{n\to\infty} \frac{\log_2^2 n}{2} = \Theta(\log^2 n). $$
using the change of indices $j = \log_2 n - i$, so that $\frac{n}{2^i} = 2^{\log_2 n - i} = 2^j$.
$(\dagger)$ Note that since we want $\Theta(\cdot)$ asymptotics, the base of the logarithm does not actually matter, but here we obtained a sharper estimate, with the tight constant.
Another method, not giving the tight constant but sufficient for your purposes: since all summands are non-negative and decreasing, $$\begin{align} \sum_{i=0}^{\log_2 n} \log_2 \frac{n}{2^i} &\geq \sum_{i=1}^{\frac{1}{2}(\log_2 n)} \log_2 \frac{n}{2^i} \geq \frac{1}{2}(\log_2 n) \cdot \log_2 \frac{n}{2^{\frac{1}{2}(\log_2 n)}} = \frac{1}{2}(\log_2 n) \cdot \log_2 \frac{n}{\sqrt{n}} \\ &=\frac{1}{2}(\log_2 n) \cdot \log_2 \sqrt{n} =\frac{1}{4}(\log_2 n) \cdot \log_2 n \\ &= \frac{1}{4} \log^2_2 n. \end{align}$$ On the other hand, we also have $$\begin{align} \sum_{i=0}^{\log_2 n} \log_2 \frac{n}{2^i} = \sum_{i=0}^{\log_2 n - 1} \log_2 \frac{n}{2^i} &\leq \log_2 n \cdot \log_2 n = \log^2_2 n \end{align}$$ since we have $\log_2 n$ non-zero terms in the sum, all less than or equal to $\log_2 \frac{n}{2^0} = \log_2 n$.