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.
A counterexample is given by
$$
f(x,y)=(x^4-y^3)^2+y .
$$
Clearly, $f(R, R^{4/3})=R^{4/3}\ll R^2 = \left( \min \{ R, R^{4/3} \} \right)^2$, so $f$ doesn't satisfy your condition.
However, $f$ is coercive: If $y\le 0$, then $f(x,y)\ge x^8+y^6 +y$ and now either $|y|$ is not large and there are no problems or if $|y|\gg 1$, then $y^6+y\ge (1/2)y^6$, say.
If $y\ge 0$, then $f(x,y)\ge \max\{ (x^4-y^3)^2, y\} \to\infty$ as $|(x,y)|\to\infty$, $y\ge 0$.
Best Answer
Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.
For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose $f$ such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to $2f(n)+1$ are
After this we will have reached $k = 6n+1$ which is smaller than $g(n)=6n+2$.
Point 2. is likely very hard and I have no idea how to tackle that. Indeed answering 1. with polynomials $f$ and $g$ with $\deg g < \deg f$ will already be very difficult. Although when taking $f$ to be the similar looking non polynomial $f=2(4^x-1)/3$ one can take $g=2$ for silly reasons.