For all $Z \in C^\wedge, Y \in D^\wedge$, we have $C^\wedge(f^\wedge Y,Z)=D^\wedge(Y,f_+ Z)$. If we put $Y = D(-,d), Z = C(-,c)$, we get

$(f_+ C(-,c))(d) = C^\wedge(f^\wedge D(-,d),C(-,c)) = C^\wedge(D(f-,d),C(-,c))$

There seems to be no connection between $f_+ C(-,c)$ and $D(-,fc)$ (only when $f$ is an equivalence). For example,

$D^\wedge(D(-,fc),f_+ C(-,c)) = C^\wedge(f^\wedge D(-,fc),C(-,c)) = C^\wedge(D(f-,fc),C(-,c))$

and it should be possible to construct an example where there is no natural transformation $D(f-,fc) \to C(-,c)$ at all.

For all $Z \in C^\wedge, Y \in D^\wedge$, we have $C^\wedge(f^\wedge Y,Z)=D^\wedge(Y,f_+ Z)$. If we put $Y = D(-,d), Z = C(-,c)$, we get

$(f_+ C(-,c))(d) = C^\wedge(f^\wedge D(-,d),C(-,c)) = C^\wedge(D(f-,d),C(-,c))$

There seems to be no connection between $f_+ C(-,c)$ and $D(-,fc)$ (only when $f$ is an equivalence). For example,

$D^\wedge(D(-,fc),f_+ C(-,c)) = C^\wedge(f^\wedge D(-,fc),C(-,c)) = C^\wedge(D(f-,fc),C(-,c))$

and it is possible to construct an example where there is no natural transformation $D(f-,fc) \to C(-,c)$ at all. For example if $D(fx,fc)$ is nonempty, but $C(x,c)$ is empty. Take $C^{op}=D=Set, f = Hom(-,2), x = 0, c = 1$.