It appears to me that the condition on $F:\cal C\to\cal D$ would be:

For any morphism $s: a\to b$ of $\cal D$, the following category is connected:

An object consists of a $\cal C$-object $c$ and a factorization $a\to F(c)\to b$ of $s$.

A morphism $c_1\to c_2$ is a $\cal C$-morphism such that the induced map $F(c_1)\to F(c_2)$ is compatible with the maps from $a$ and to $b$.

I don't recall ever having run into this sort of 'two-sided comma category' before, but it seems to be the answer.

I got this by choosing $G:\cal D$ to be represented by the object $a$ and thinking about the fiber of the map $\epsilon: (F_!F^*G)(b)\to G(b)$ over the element $s$.

It appears to me that the condition on $F:\cal C\to\cal D$ would be:

For any morphism $s: a\to b$ of $\cal D$, the following category is connected:

An object consists of a $\cal C$-object $c$ and a factorization $a\to F(c)\to b$ of $s$.

A morphism $c_1\to c_2$ is a $\cal C$-morphism such that the induced map $F(c_1)\to F(c_2)$ is compatible with the maps from $a$ and to $b$.

I don't recall ever having run into this sort of 'two-sided comma category' before, but it seems to be the answer.

I got this by choosing $G$ from $\cal D$ to Set to be represented by the object $a$ and thinking about the fiber of the map $\epsilon: (F_!F^*G)(b)\to G(b)$ over the element $s$.