You're right; this is an error in what you've heard. The embedding $i:P\to \hat P$ of a poset $P$ into its complete poset of downsets $\hat P$ is the free completion of the poset, in the sense that every poset morphism
$P\to Q$, where $Q$ is complete, has a unique factorization through $i$ preserving all joins. However, $[-\infty,\infty]$ is not quite the free completion of $\mathbb{Q}$. Indeed, there is no factorization of $i:\mathbb Q\to \widehat{\mathbb Q}$ through the inclusion $\mathbb{Q}\to [-\infty,\infty]$ preserving joins, since the join of the set $\{1-1/n:n\in\mathbb N\}$ in $[-\infty,\infty]$ is $1$, while in $\widehat{\mathbb Q}$ it is $\{x:x<1\}$.
The property that $[-\infty,\infty]$ does have, and more generally that characterizes Dedekind-MacNeille completions $j:P\to \bar P$ of posets, is that $[-\infty,\infty]$ is the universal complete poset admitting a map from $\mathbb{Q}$ which preserves joins which already exist. This is in contrast to $i:\mathbb Q\to \widehat{\mathbb Q}$, which destroys the join mentioned at the end of the last paragraph.
Indeed free cocompletions destroy whatever colimits in a given enriched setting are not absolute-joints of finite sets in posets, splitting of idempotents in ordinary categories, biproducts in categories enriched over abelian groups, etc. So if $Q$ is some complete poset, then maps $\mathbb Q\to Q$ which preserve joins factor uniquely through $[-\infty,\infty]$. We could say that $[-\infty,\infty]$ is the free conservative cocompletion of $\mathbb{Q}$-the cocompletion that doesn't mess up anything we've already handled.
Arguably the Dedekind completion is a bit more abstruse, categorically, than the Cauchy completion. If one views $\mathbb{Q}$ as a category enriched over $[0,\infty]$, that is, as a generalized metric space, then the Cauchy completion is just the closure of $\mathbb{Q}$ in its category of enriched presheaves under absolute colimits-the absolute colimits in metric spaces being essentially limits of Cauchy sequences. For this reason, the closures of other kinds of enriched categories under absolute colimits are sometimes also called Cauchy completions.
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
Let: $$\Gamma = \operatorname{colim} \left\{ \left(\int F\right)^{op} \to C \to [C^{op}, \mathbf{Set}] \right\}.$$
Then a morphism $u : \Gamma \to Z$ is the same thing as a collection of morphisms: $$u_{c,x} : Y(c) \to Z$$ for all $c \in C$ and $x \in F(c)$, such that if $\varphi : d \to c$ is a morphism in $C$, then the obvious commutative diagram commutes. More precisely, we should have: $$u_{d, F(\varphi)(x)} = u_{c,x} \circ Y(\varphi). \tag{1}$$
This collection is the same thing as a collection $U_c : Y(c) \times F(c) \to Z$ for $c \in C$, defined by: $$U_c(\xi, x) = u_{c,x}(\xi)$$ for all $c \in C$, $x \in F(c)$ and $\xi \in Y(c)$. If you inspect what the equation $(1)$ means for $U$, you will see that it exactly means that $U$ can factor through the coequalizer that usually defines the coend: $$\int^{c \in C} Y(c) \times F(c) = \operatorname{coeq} \left\{ \bigsqcup_{c,d} Y(c) \times F(d) \rightrightarrows \bigsqcup_c Y(c) \times F(c) \right\}$$ And we can check that this defines a bijection on the level of $\hom$-sets (the reverse construction, given some $U$, simply sets $u_{c,x}(\xi) = U_c(\xi,x)$): $$\hom(\Gamma, Z) \cong \hom(\int^c Y(c) \times F(c), Z).$$ I'll let you check that it's natural in $Z$ (simple exercise in definitions), so that finally: $$\Gamma \cong \int^c Y(c) \times F(c).$$