Here's one possible answer to this question.
Let's take the viewpoint that functors are representations of categories.
First, why is this sensible?
Well, recall that categories are generalizations of monoids (and consequently groups as well), since a one object category is the same thing as a monoid.
If $M$ is a monoid, then we can define a category, $C$, with one object, $*$, hom set $C(*,*)=M$, and unit and composition given by the unit and multiplication in $M$. Conversely, given a one object category $C$, $C(*,*)$ is a monoid with composition as multiplication, and these constructions are inverse to each other.
From now on, if $M$ is a monoid, or $G$ is a group, I'll write $BM$ or $BG$ for the corresponding one object category.
Now, what about functors? Well, what are functors $[BG,k\newcommand\Vect{\text{-}\mathbf{Vect}}\Vect]$?
Well, we need to pick a vector space $V$ to send $*$ to, and we need to pick a monoid homomorphism $G\to \newcommand\End{\operatorname{End}}\End V$. Since $G$ is a group, this is equivalent to a group homomorphism $G\to \operatorname{GL}(V)$. In other words, functors from $BG$ to $k\Vect$ are exactly the same as linear group representations, and you can check that natural transformations of functors correspond exactly to the $G$-equivariant linear maps.
Similarly, when we replace $k\Vect$ with $\newcommand\Ab{\mathbf{Ab}}\Ab$, or $\newcommand\Set{\mathbf{Set}}\Set$, we get $G$-modules and $G$-sets respectively.
Specifically, these are all left $G$-actions, since a functor $F:BG\to \Set$ must preserve composition, so
$F(gh)=F(g)F(h)$, and we define $g\cdot x$ by $F(g)(x)$. Thus $(gh)\cdot x = g\cdot (h\cdot x))$.
A contravariant functor $\newcommand\op{\text{op}}BG^\op\to \Set$ gives a right $G$-action, since now
$F(gh)=F(h)F(g)$, so if we define $x\cdot g = F(g)(x)$, then we have
$$x\cdot (gh) =F(gh)(x) = F(h)F(g)x = F(h)(x\cdot g) = (x\cdot g)\cdot h.$$
Thus we should think of covariant functors $[C,\Set]$ as left $C$-actions in $\Set$,
and we should think of contravariant functors $[C^\op,\Set]$ as right $C$-actions in $\Set$.
Yoneda Lemma in Context
Representable presheaves now correspond to free objects in a single variable in the following sense.
The Yoneda lemma is that we have a natural isomorphism
$$
[C^\op,\Set](C(-,A),F)\simeq F(A)\simeq \Set(*,F(A)).
$$
In other words, $C(-,A)$ looks a lot like the left adjoint to the "forgetful" functor that sends a presheaf $F$ to its evaluation at $A$, $F(A)$, but evaluated on the singleton set $*$.
In fact, we can turn $C(-,A)$ into a full left adjoint by noting that
$$\Set(S,F(A)) \simeq \prod_{s\in S} F(A) \simeq \prod_{s\in S}[C^\op,\Set](C(-,A),F)
\simeq [C^\op,\Set](\coprod_{s\in S} C(-,A), F),$$
and $\coprod_{s\in S} C(-,A)\simeq S\times C(-,A)$.
Thus one way of stating the Yoneda lemma is that $S\mapsto S\times C(-,A)$ is left adjoint to the evaluation at $A$ functor (in the sense that the two statements are equivalent via a short proof). Incidentally, there is also a right adjoint to the evaluation at $A$ functor, see here for the argument.
Relating this back to more familiar notions
First thing to notice in this viewpoint is that we now have notions of "free on an object" rather than just "free." I.e., I tend to think of $C(-,A)$ as being the free presheaf in one variable on $A$ (this is not standard terminology, just how I think of it).
Now we should be careful, a free object isn't just an object, it's an object and a basis. In this case, our basis (element that freely generates the presheaf) is the identity element $1_A$.
Thinking about it this way, the proof of the Yoneda lemma should hopefully be more intuitive. After all, the proof of the Yoneda lemma is the following:
$C(-,A)$ is generated by $1_A$, since $f^*1_A=f$, for any $f\in C(B,A)$, so natural transformations $C(-,A)$ to $F$ are uniquely determined by where they send $1_A$. (Analogous to saying $1_A$ spans $C(-,A)$). Moreover, any choice $\alpha\in F(A)$ of where to send $1_A$ is valid, since we can define a natural transformation by "extending linearly" $f=f^*1_A \mapsto f^*\alpha$ (this is analogous to saying $1_A$ is linearly independent, or forms a basis).
The covariant version of the Yoneda lemma is the exact same idea, except that we are now working with left representations of our category.
Examples of the Yoneda lemma in more familiar contexts
Consider the one object category $BG$, then the Yoneda lemma says that
the right regular representation of $G$ is the free right $G$-set in one variable (with the basis element being the identity, $1_G$).
(The free one in $n$-variables is the disjoint union of $n$ copies of the right regular representation.)
The embedding statement is now that $G$ can be embedded into $\operatorname{Sym}(G)$ via $g\mapsto -\cdot g$.
This also works in enriched contexts. A ring is precisely a one object category enriched in abelian groups, and the Yoneda lemma in this context says that the right action of $R$ on itself (often denoted $R_R$) is the free right $R$-module in one variable, with the basis being the unit element $1_R$.
(The free one in $n$-variables is now the direct sum of $n$ copies of $R_R$)
The embedding statement here is that $R$ can be embedded into the endomorphism ring of its underlying abelian group via $r\mapsto (-\cdot r)$.
Best Answer
This is defenitly not an awnser, since it is full of vagueries, half truths, and bad analogies. But the comment section was to small to fit this remarks.
There is not always a functor $C\to D^{C^{op}}$, for this is equivalent as having a functor $C^{op}\times C \to D$. Enriched categories over $D$ provide such a functor. A category is a $\text{Set}$-enriched category, the functor $C^{op}\times C \to \text{Set}$ is given by the $\text{Hom}$ sets of your category $C$. Instead you always have a natural functor in $C$, $C \to D^{(D^C)}$ which takes an object $x$ of $C$ to the evaluation at $x$.
Note that the Yoneda lemma is very often stated as the fact that the evaluation functor at $x$ is represented by the functor represented by $x$, and then the functor $C \to \text{Set}^{(\text{Set}^C)}$ is given by composing $C \to (Set^{C})^{op}$ with $(Set^{C})^{op} \to \text{Set}^{(\text{Set}^C)}$
This parallels quite faithfully what happens in $k$-vector spaces : if you have an inner product on a vector space $V$, i.e. a map $V \otimes V \to k$, you get a linear map $V \to V^*$, and having such a linear map gives you an inner product. An inner product is not degenerated if this map is injective, giving some kind of Yoneda vibe to the situation.
Replacing $k$ with $\text{Set}$, you can think of categories as some kind of 'modules' over the category $\text{Set}$, having an 'inner product' (the $\text{Hom}$ spaces) that is 'not degenerated' (Yoneda embedding).
But you always have a natural linear map $V \to V^{**}$ that sends a vector $v$ to the linear map which evaluates a linear form on $V$ at the vector $v$.