If I've understood your first question correctly then you can take $g$ and $h$ to be the sum of monomials corresponding to the vertices of $Q$ and $R$ respectively. Then $\tilde{f}=gh$ does the job.

For your follow up question, the answer is negative. Consider $S_g=\{(0,0),(1,0)\}$ and $S_h=\{(0,0),(0,1),(0,2)\}$ and $S_f=\{(0,0),(1,0),(0,2),(1,2)\}$. Since $g=a+bx$ and $h=c+dy+ey^2$ for non-zero $a,b,c,d,e$ the coefficient of $y$ in $gh$ is $ad\neq 0$ but $(0,1)\notin S_f$.

The answer to the stronger question in the edit is also negative. Consider $$P=\{(0,0),(0,1),(0,2),(1,0),(1,2),(2,0),(2,1),(2,2)\}$$ which is decomposable for example by taking $Q=\{(0,0),(1,0),(2,0)\}$ and $R=\{(0,0),(0,1),(0,2)\}$. The idea is that this is the only choice of $Q,R$ that can decompose $P$ and this is not hard to check by hand. However $P$ is missing $(1,1)$ from $Q+R$ so we cant have $P=Q+R$. This example works over any field.