Suppose $f,g \in k[x,y]$ and $I:=\left<f,g\right>$ is an ideal.
My question is if the second implication is true.
$I$ is zero dimensional $\implies$ $I \cap k[x] \neq \left<0\right> $ and $I\cap k[y] \neq \left<0\right>$ $\implies$ $ f \in I \cap k[x] $ and $ g \in I \cap k[y] .$
The implication is definitely false in general. A more interesting question would be, whether one can always replace the two generators by elements in $k[x]$ and $k[y]$ respectively. The answer is also negative.
Consider $I=(x^2,x+y)$. Note that $I \cap k[x]=(x^2)$ and $I \cap k[y]=(y^2)$. If $I=(f,g)$ with $f \in k[x], g \in k[y]$ we would have $x^2|f$ and $y^2|g$, in particular $(f,g) \subset (x^2,y^2)$ and thus $x+y \notin (f,g)$, contradiction!