Let $f_1,g_1,f_2,g_2 \in k[t]$, $k$ a field of characteristic zero.
(Edit: Assume that $k(f_1,g_1)=k(t)$ and $k(f_2,g_2)=k(t)$).
By this question, there exist $a_1,b_1 \in k$ such that $\deg(\gcd (f_1-a_1,g_1-b_1)) =1$, and there exist $a_2,b_2 \in k$ such that $\deg (\gcd (f_2-a_2,g_2-b_2)) =1$.
Remark: I do not care about $c_1,c_2 \in k$ such that $\gcd (f_1-a_1,g_1-b_1) = t-c_1$ and $\gcd (f_2-a_2,g_2-b_2) = t-c_2$. (Namely, it is fine with me if $c_1 \neq c_2$).
Is it possible to find $a,b \in k$ such that $\deg (\gcd (f_1-a,g_1-b)) =1$, and
$\deg (\gcd (f_2-a,g_2-b)) =1$?
From what I understand, we can take $a:=a_1=a_2$, but I am not sure if we can find the same $b$.
A relevant comment can be found here = the last comment of Gerry Myerson: "... I think it says for (almost) every $c$ there exist $a,b$ such that $\gcd(f−a,g−b)=t−c$. So for (almost) every $a$ there exist $b,c$ such that $\gcd(f−a,g−b)=t−c$". The problem is that, after fixing a 'good' $a$ for both $(f_1,g_1)$ and $(f_2,g_2)$, I am not sure if we can find 'good' $b_1=b_2$.
Even if there exist infinitely many 'good' pairs for $(f_1,g_1)$, $U:=\{(a_1,b_1)| \deg(\gcd(f_1-a_1,g_1-b_1))=1\}$, and infinitely many 'good' pairs for $(f_2,g_2)$, $V:=\{(a_2,b_2)| \deg(\gcd(f_2-a_2,g_2-b_2))=1\}$, I guess that there is no guarantee that $U \cap V$ is nonempty.
Any hints are welcome!
What does $\deg(\gcd(f-a,g-b))=1$ mean? It means $\gcd(f-a,g-b)=t-c$ for some $c$ in $k$, which implies that $f(c)=a$ and $g(c)=b$.
So, for your question, you want $c_1,c_2$ in $k$ such that $f_1(c_1)=f_2(c_2)=a$ and $g_1(c_1)=g_2(c_2)=b$.
But given polynomials $f_1,f_2$ there's no reason why there should exist $c_1,c_2$ such that $f_1(c_1)=f_2(c_2)$. E.g., over the reals, if $f_1(t)=t^2+1$ and $f_2(t)=-t^2-1$ then the ranges of $f_1$ and $f_2$ are disjoint.