Here is a proof of the generalization of your Weak conjecture to the ring $\mathbf{Z}/m\mathbf{Z}$ where $m$ is any odd positive integer. First let me clarify what is being proved. Let $S$ be the subgroup of $(\mathbf{Z}/m\mathbf{Z})^\times$ which is generated by $2$, and consider a directed graph whose vertices are the sets $gS$ with $g\in\mathbf{Z}/m\mathbf{Z}$ (I emphasize that $g$ need not be in $(\mathbf{Z}/m\mathbf{Z})^\times$), and which has a directed edge from $gS$ to $(3g+1)S$ for every $g\in\mathbf{Z}/m\mathbf{Z}$. I will show:
1) If $3\nmid m$ then this directed graph is strongly connected (in the sense that there is a directed path from any vertex to any other vertex).
2) If $3\mid m$ then for any $g,h\in(\mathbf{Z}/m\mathbf{Z})^\times$ there is a directed path from the vertex $gS$ to the vertex $hS$ (although this path might go through vertices of the form $iS$ with $i\notin (\mathbf{Z}/m\mathbf{Z})^\times$).
Both of these are special cases of the Theorem proved below.
Lemma. Let $n$ be a positive integer, and let $M$ be a subset of $\mathbf{Z}/n\mathbf{Z}$ which contains elements $u,v$ for which $u-v\in(\mathbf{Z}/n\mathbf{Z})^\times$. Then, for every sufficiently large positive integer $s$, every element of $\mathbf{Z}/n\mathbf{Z}$ is the sum of exactly $s$ elements of $M$.
Proof. Let $M_r$ be the set of all sums of exactly $r$ elements of $M$, so that $M_1=M$. Pick $u,v\in M_r$ for which $u-v\in(\mathbf{Z}/n\mathbf{Z})^\times$. Plainly $M_{r+1}$ contains $M_r+u$ and $M_r+v$, so that $\#M_{r+1}\ge\#M_r$. If $\#M_{r+1}=\#M_r$ then we must have $M_r+u=M_r+v$, or equivalently $M_r+(u-v)=M_r$. But then $M_r+\ell(u-v)=M_r$ for any positive integer $\ell$, whence $M_r+w=M_r$ for any $w\in\mathbf{Z}/n\mathbf{Z}$, so we must have $M_r=\mathbf{Z}/n\mathbf{Z}$. Thus, for every $r>0$, either $\#M_{r+1}>\#M_r$ or $M_r=\mathbf{Z}/n\mathbf{Z}$. Since $\#M_r\le m$ for every $r$, it follows that there is some $r$ for which $M_r=\mathbf{Z}/n\mathbf{Z}$, whence $M_s=\mathbf{Z}/n\mathbf{Z}$ for every $s\ge r$. This finishes the proof of the Lemma.
Theorem. Write $m=3^kn$ where $k\ge 0$ and $3\nmid n$. Pick any $g,h\in\mathbf{Z}/m\mathbf{Z}$, and suppose that either $k=0$ or $3\nmid h$.
Then there is a directed path from $g\langle 2\rangle$ to $h\langle 2\rangle$.
Proof. Apply the Lemma with $M$ being the subgroup of $(\mathbf{Z}/n\mathbf{Z})^\times$ generated by $2$ (and with $u=2$ and $v=1$) to obtain a corresponding $s$. Let $r$ be the order of $3$ in $(\mathbf{Z}/n\mathbf{Z})^\times$. Let $\ell$ be the smallest integer such that $\ell>k/r$. Recall that every element of $U:=(\mathbf{Z}/3^k\mathbf{Z})^\times$ is a power of $2$. Since an element $i\in \mathbf{Z}/3^k\mathbf{Z}$ lies in $U$ if and only if $i+3/2$ lies in $U$, it follows that every element of $U$ can be written as $-3/2+2^t$ for some positive integer $t$. The hypothesis that either $k=0$ or $3\nmid h$ implies that the image of $h$ in $\mathbf{Z}/3^k\mathbf{Z}$ is actually in $U$, so there is some $t>0$ for which $h\equiv -3/2+2^t\pmod{3^k}$. Let $b_0,\dots,b_{s-1}$ be nonnegative integers for which $h-3g+s-2^t\equiv\sum_{i=0}^{s-1} 2^{b_i}\pmod{n}$. Define $$a_0:=t,$$
$$a_i:=0 \,\,\text{if either $0<i<r\ell$ or $r\nmid i$},$$
$$a_{r(\ell+j)}:=b_j \,\,\text{for $0\le j\le s-1$}.$$
Then
$$x:=3^{r(\ell+s-1)+1}2^0g+\sum_{i=0}^{r(\ell+s-1)} 3^i 2^{a_i}$$
equals
$$3^{r(\ell+s-1)+1}g + \sum_{i=0}^{r(\ell+s-1)} 3^i + \sum_{j=0}^{s-1} 3^{r(\ell+j)} (2^{b_j}-1) + (2^t-1)$$
$$=3^{r(\ell+s-1)+1}g + \frac{3^{r(\ell+s-1)+1}-1}{3-1} + \sum_{j=0}^{s-1} 3^{r(\ell+j)} (2^{b_j}-1) + (2^t-1).$$
Here $x$ is congruent mod $3^k$ to
$$-\frac12+(2^t-1)=2^t-\frac32=h.$$
Next, $x$ is congruent mod $n$ to
$$3g+\frac{3-1}{3-1}+\sum_{j=0}^{s-1} (2^{b_j}-1) + (2^t-1)=
3g+1+(h-3g+s-2^t)-s+(2^t-1)=h.$$
Therefore $x=h$.
An easy induction shows that the union of the sets $v\langle 2\rangle$ for which the vertex $v\langle 2\rangle$ can be reached from $g\langle 2\rangle$ via a directed path of length $z$ is $\{3^z2^{d_z}g +\sum_{i=0}^{z-1} 3^i 2^{d_i}\colon d_i\in\mathbf{Z}\}$. Thus there is a directed path of length $r(\ell+s-1)+1$ from $g\langle 2\rangle$ to $h\langle 2\rangle$. This finishes the proof of the Theorem.
What is $k_n$ (the ones you checked for $n \le 10^8$)?
Some thought on the champions, but I didn't dig much yet:
I think the tree must be symmetric up to $[k-1]$ (I'll come to symmetry later), you could still have $c1=c2$ without symetry, but it would probably not hold with the other rules (e.g. alternance). I have to check.
For visual purpose, I put the "$\{2n\}$" leafs on the left in the left part of the tree, and on the right in the right part of the tree. I put multiple of $3$ in red, and circled the "$n \equiv 2 \pmod 3$".
I call "symmetric counterparts" elements which map to each other in the symmetry (e.g. $69$ and $213$ are symmetric counterparts. $80$ and $26$ are symmetric counterparts). For the symmetry to be kept, every counterparts must be of the same value modulo $3$ ($0$, $1$ or $2$).
From the example above, you can see that the symmetry is lost on the $2^{k-2}-1=7$ element, which is not $\equiv 2 \pmod 3$ like its counterpart $23$, so it has one child less.
It is well known that in a Collatz sub-sequence starting with $2^a-1$, you will always climb up to $3^a-1$ in $a$ steps, which is the largest "forward sub-sequence" up to this $n=2^a-1$, so my guess is that having a champion has something to do with that (especially knowing that the forward sub-sequence must have a symmetric counterpart of the same kind, which is the case as shown bellow).
So for the example above, on the left of the tree we have a forward sub-sequence starting with $2^{k-1}-1=15$: $\{15,23,35,53,80\}$, ending with $3^{k-1}-1=80$, and on the right, the (almost) symmetric counterpart sub-sequence starting with $2^{k-2}-1=7$: $\{7,11,17,26\}$, ending with $3^{k-2}-1=26$.
Note that $[k-1]$ is even since $[k+1]=\alpha (n)$ is even, so $3^{k-1}-1$ can be rewritten $9^{\frac{k-1}{2}}-1$, and from there you divide by $4$ to reach $n=20$ which leads to your formula $n=\frac{9^{\frac{k-1}{2}}-1}{4}$. The $\{(2n-1)/3\}$ element ($13$ in the example) is therefore $\frac{(2\cdot\frac{9^{\frac{k-1}{2}}-1}{4})-1}{3}=\frac{(\frac{3^{k-1}-1}{2})-1}{3}=\frac{3^{k-2}-1}{2}$ and the parent ($26$) is therefore $3^{k-2}-1$ which we know leads up to $2^{k-2}-1=7$ but in 1 step less than its counterpart, which gives a symmetry break ($2^a-1$ is not $\equiv 2 \pmod 3$ like its successors).
For the symmetry part, I still have to look, but perhaps this kind of thing might do the trick:mod 18 (probably depends on $k$)
I also have to look at other possible "symmetry breaks", and probably draw some other cases to find some rules as soon as I have more time.
Best Answer
This problem is still open. See for example Sections 2.6 and 2.7 of Lagarias's survey.