The second sum, can be rewritted as
$$\sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{j=\max\{n,k^{1/\alpha}\}\atop k\mid j}^{n^2} \frac{1}{j^\lambda} = \sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{\ell=\lceil \max\{n/k,k^{1/\alpha-1}\}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{(k\ell)^\lambda} = \sum_{k=1}^{n^{2\alpha}} \sum_{\ell=\lceil \max\{n/k,k^{1/\alpha-1}\}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda}$$
$$=\sum_{k=1}^{n^{\alpha}}\sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} + \sum_{k=n^{\alpha}}^{n^{2\alpha}} \sum_{\ell=\lceil k^{1/\alpha-1}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda}.$$
The inner sums can be bounded by the integrals:
$$\int_{L+1}^{n^2/k} \frac{d\,t}{t^\lambda} \leq \sum_{\ell=\lceil L\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq \int_{L-1}^{n^2/k} \frac{d\,t}{t^\lambda}.$$
E.g., for $\lambda>1$, we have
$$C_1k^{\lambda-1} \leq \sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq C_2k^{\lambda-1},$$
where
$$C_1=\frac{n^{\lambda-1}-1}{(\lambda-1)n^{2(\lambda-1)}}, \qquad C_2= \frac{(n-n^\alpha)^{1-\lambda}-1}{(\lambda-1)n^{2(\lambda-1)}}.$$
Then
$$\sum_{k=1}^{n^{\alpha}} \sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \geq \sum_{k=1}^{n^{\alpha}} C_1 k^{\lambda-1}\geq \frac{C_1}{\lambda} n^{\alpha\lambda}$$
and
$$\sum_{k=1}^{n^{\alpha}} \sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq \sum_{k=1}^{n^{\alpha}} C_2 k^{\lambda-1}\leq \frac{C_2}{\lambda} (n^\alpha+1)^{\lambda}.$$
The other sum is estimated similarly.