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.