This is a bit too long to be a comment, but hopefully it can shed light on your question.
Adjointness via representability
The concept of adjointness, like many others in category theory, can be understood in terms of representable functors.
For any functor $\mathcal{C} \xrightarrow{F} \mathcal{D}$, we can define a functor \begin{alignat}{2} \mathcal{D} &\xrightarrow{G^{\mathsf{formal}}} &&[\mathcal{C}^{\mathop{op}}, \mathbf{Set}] \\ d &\longmapsto &&(c \mapsto \mathcal{D}(F c, d)). \end{alignat}
This looks a bit like the Yoneda embedding of $\mathcal{D}$, except its target is the presheaf on $\mathcal{C}$ which first applies $F$ and then takes the Hom functor in $\mathcal{D}$ fixed in its second argument.
Now we can claim the following.
Proposition. $F$ has a right adjoint if and only if $G^{\mathsf{formal}} (d)$ is representable for every object $d$ of $\mathcal{D}$.
For the forwards direction, assuming that $F \dashv G$, we get the representability of any functor in the image of $G^{\mathsf{formal}}$ by the natural bijection $\mathcal{D} (F c, d) \cong \mathcal{C} (c, G d)$. For the reverse direction, define the right adjoint to $F$ to be $Y^{-1} \circ G^{\mathsf{formal}}$ where $Y^{-1}$ is the inverse to the Yoneda embedding defined only on the representable functors $[\mathcal{C}^{\mathop{op}}, \mathbf{Set}]$ (every such one is in the essential image of $Y$). Then deduce $$ \mathcal{C}(c, Y^{-1} G^{\mathsf{formal}} d) \cong [\mathcal{C}^{\mathop{op}}, \mathbf{Set}](Y c, G^{\mathsf{formal}} d) \cong G^{\mathsf{formal}} d (c) = \mathcal{D} (F c, d), $$ naturally in $c$ and $d$, where the first step uses the adjoint equivalence $Y \dashv Y^{-1}$ and the second step is the Yoneda lemma.
This means that the right adjoint $G$ is a universal solution to the equation $\mathcal{D} \xrightarrow{G} \mathcal{C} \xrightarrow{Y} [\mathcal{C}^{\mathop{op}}, \mathbf{Set}] \cong \mathcal{D} \xrightarrow{G^{\mathsf{formal}}} [\mathcal{C}^{\mathop{op}}, \mathbf{Set}]$. The formal right adjoint to $F$, $G^{\mathsf{formal}}$, always exists, and it tells you how to build the $G$ under certain conditions.
A tangent on profunctors
If you like, you can see $G^{\mathsf{formal}}$ as a profunctor by uncurrying, and this leads to the (slightly disingenuous) slogan:
Every functor has a right adjoint, but it is a profunctor instead of a functor.
Now, up to Cauchy-completeness, every profunctor which admits a right adjoint in $\mathbf{Prof}$ is equivalent to a functor; analogously, the profunctor obtained from $G^{\mathsf{formal}}$ is equivalent to some functor $G$ exactly when it is a right-adjoint profunctor. This leads to an enhanced slogan:
Every functor has a right adjoint, but sometimes it is profunctor instead of a functor.
In this case, 'sometimes' means that it does not have a right adjoint in the usual sense.
