Collatz, Sort, Repeat Background
The Collatz (or 3x+1) map (A006370) is defined as the following:
$$
  a(n) =
\begin{cases}
n/2,  &amp; \text{if $n$ is even} \\
3n+1, &amp; \text{if $n$ is odd}
\end{cases}
$$
Now, let's define the sequence \$s_0\$ as the infinite sequence of positive integers \$1, 2, 3, 4, \cdots\$, and define \$s_k\$ as the result of applying the process \$k\$ times to \$s_0\$:

Apply Collatz map to each term of the input sequence.
Sort all the terms to get a non-decreasing sequence.

Here are first few terms of \$s_0, \cdots, s_3\$:
s0: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, ...
s1: 1, 2, 3, 4, 4, 5, 6, 7, 8, 9, 10, 10, 11, 12, 13, 14, 15, 16, 16, 17, ...
s2: 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 8, 8, 9, 10, 10, 11, 11, 12, 13, 14, 14, ...
s3: 1, 1, 2, 2, 3, 4, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 10, 10, 10, 11, 11, 12, ...

Challenge
Given \$k\$ as input, output the sequence \$s_k\$.
Default sequence I/O methods are allowed. For this challenge, this means you can use one of the following:

Take \$k\$ as input, and output the terms of \$s_k\$ forever.
Take \$k, n\$ as input, and output the value of \$s_k(n)\$ (0- or 1-indexed).
Take \$k, n\$ as input, and output the first \$n\$ values of \$s_k\$.

You can also use 0- or 1-indexing for \$k\$.
Standard code-golf rules apply. The shortest code in bytes wins.
Python 2, 81 bytes

f=lambda k,n:k and sorted([x/2,x*3+1][x%2]for x in f(k-1,n*2))[:n]or range(1,n+1)

Try it online!
-19 bytes thanks to Leo and Bubbler
-11 bytes thanks to dingledooper (-6 + -5 from Python 2)
-8 bytes by swapping over to the brute force solution
-2 bytes thanks to Noodle9
To get \$s_k\$ up to \$n\$ elements, we only need to get \$s_{k-1}\$ up to \$2n\$ elements.