Naturality can be detected on a dense subcategory $\require{AMScd}\def\colim{\text{colim}}$I need this result in less generality, but I'd be happy to know this stronger version holds.
Let $\{\alpha_c : Fc \to Gc\}$ be arrows in a category $D$, indexed by the objects of a category $C$, for two functors $F,G: C\to D$. 
Let $A\subseteq C$ be a dense subcategory (meaning that $i : A \hookrightarrow C$ is a dense functor). Then $\alpha$ is a natural transformation if and only if it is natural on the components $c\in A$ (and with respect to morphisms of $A$).
Edit: This is not what I wanted to know (and false, see below): instead, I'm trying to prove that there is a unique extension of a natural transformation $Fi\to Gi$ to a natural transformation $F\to G$.
Example: for a functor $K: [I°,Set] \to [I°,Set]$, a natural transformation $\alpha : 1_{[I°,Set]}\to K$ is such if and only if it is natural when restricted to representables.
It is easy to see that the right universal property gives that if $P = \colim y(X_i)$ is a colimit of representables, then there is
$$
P \cong \colim\; y(X_i) \xrightarrow{\colim \alpha_{X_i}} \colim\; Ky(X_i) \to K\big(\colim\; y(X_i)\big) \cong KP
$$
It seems then that all boils down to the fact that the morphism $\colim\; KX\to K(\colim\; X)$ is "natural", i.e.
$$
\begin{CD}
KX @>>> K(\colim_I X_i)\\
@VVV @VVV\\
KY @>>> K(\colim_J Y_j)
\end{CD}
$$
commutes (but how do you induce the right-vertical arrow?).
 A: $$
\newcommand{\colim}{\mathop{\rm colim}\nolimits}
\newcommand{\Nat}{\mathop{\rm Nat}\nolimits}
\newcommand{\cancom}[1]{\big((i/#1)\stackrel{\mathrm{pr}_}{→}  \stackrel{i}{→} ℂ\big)}
$$
First, I thank Ohad Kammar, Mathieu Huot and Sean Moss who helped me to come up with this answer.
To answer your EDIT in a nutshell: the result doesn't stand in general when $F$ is not cocontinuous.
Here are two counter-examples to the existence and unicity of your natural transformation extension:
Counter-example to existence
Let


*

*$ℂ =  \, ≝ \, X ⟶ Y$ (category with two objects and one non-trivial morphism between them)

*$ \, ≝ \, \mathbb{1}$ (terminal category)

*$i: \begin{cases}
    ↪︎ ℂ  \\
   \ast ⟼ Y
\end{cases}$

*$F \, ≝ \, \mathrm{const}_Y ∈ [ℂ, ]$ (constant functor sending all the $ℂ$-objects to $Y$)

*$G \, ≝ \, \mathrm{Id}_{ℂ} ∈ [ℂ, ]$


then there is a natural transformation between $Fi: \begin{cases}
   \mathbb{1} ⟶  \\
   \ast ⟼ Y
\end{cases}$ and $Gi: \begin{cases}
   \mathbb{1} ⟶  \\
   \ast ⟼ Y
\end{cases}$ (the identity), but no natural transformation from $F$ to $G$, as there is no morphism from $F(X) = Y$ and $G(X) = X$.
Counter-example to uniqueness


*

*As before, let's set $ℂ ≝ \, X ⟶ Y, \;  \, ≝ \, \mathbb{1}, \; i: \begin{cases}
    ↪︎ ℂ  \\
   \ast ⟼ Y
\end{cases}$

*Let $$ be the category given by


*

*two object $X'$ and $Y'$

*two non-trivial morphisms: $g: X' ⟶ Y'$ and $f: X' ⟶ X'$ such that $$gf = g \qquad \text{ and} \qquad f^2 = \mathrm{id}_{X'}$$


*$F = G: \begin{cases}
  ℂ ⟶  \\
  X ⟼ X' \\
  Y ⟼ Y'
\end{cases}$


then there is a natural transformation between $Fi: \begin{cases}
   \mathbb{1} ⟶  \\
   \ast ⟼ Y'
\end{cases}$ and $Gi: \begin{cases}
   \mathbb{1} ⟶  \\
   \ast ⟼ Y'
\end{cases}$ (the identity), but two distinct natural transformations from $F$ to $G$ (the identity and the one whose component at $X$ is $f$).
Fixed statement
Your assertion is true if $F$ is assumed to be cocontinuous though:

Let $F: ℂ ⟶ $ be a cocontinuous functor, $G: ℂ ⟶ $ a functor and $ \overset{i}{\hookrightarrow} ℂ$ a dense subcategory. There is a unique extension of every natural transformation $Fi ⇒ Gi$ to a natural transformation $F ⇒ G$.

The proof is as shown in the diagram below: for every $c \cong \colim\cancom{c}, \, c' \cong \colim\cancom{c} ∈ ℂ, \, f: c ⟶ c'$:


*

*there is a cocone from $\cancom{c}\stackrel{F}{→} $ to $G(c)$ (resp. from $\cancom{c'}\stackrel{F}{→} $ to $G(c')$)


*

*indeed: the green subdiagram commutes (since $Gϕ_1 = G ϕ_2 \circ Gg$ (cocone property) and $Gg \circ α_{ia_1} = α_{ia_2} \circ Fg$ (naturality))


so there exists a unique corresponding cocone morphism $α_c: F(c) ⟶ G(c)$ (resp. $α_c': F(c') ⟶ G(c')$), due to $F$ being concontinuous

*the two cocones $\cancom{c}\stackrel{F}{→} $ to $G(c)$ obtained by post-composing by $α_c \circ Ff$ and $Gf \circ α_{c'}$ are equal


*

*indeed: the red subdiagram commutes since $Gϕ_1' \circ α_{ia_1'} = α_{c'} \circ F ϕ_1'$ (colimit property of $F(c')$) and $α_c \circ Ff \circ Fϕ_1' = Gf \circ Gϕ_1' \circ α_{ia_1'}$ (colimit property of $F(c)$))


so there is a unique cocone morphism from $\cancom{c}\stackrel{F}{→} $ to $G(c)$ for these cocones, and the red square commutes: $Gf \circ α_{c'} = α_c \circ Ff$, which ends the proof.

Link to the svg picture
By duality, one can also prove that:

Let $F: ℂ ⟶ $ be a functor, $G: ℂ ⟶ $ a continuous functor and $ \overset{i}{\hookrightarrow} ℂ$ a co-dense subcategory. There is a unique extension of every natural transformation $Fi ⇒ Gi$ to a natural transformation $F ⇒ G$.


Link to the svg picture
Application: Nerve functor fully faithful
With the dual version of the lemma, we get this well-known result as a corollary:

Corollary: If $ℂ \stackrel{i}{\hookrightarrow} $ is dense, the nerve functor
  $$N_i: \begin{cases}
   &⟶ \widehat ℂ  \\
  d &⟼ (i(-), d)
\end{cases}$$
  is fully faithful.


Link to the svg picture
Proof:  We want to prove that for all $d, d' ∈ $:
$$\Nat\Big(\underbrace{(i(-), d)}_{\mathbb{y}_(d)i}, \, \underbrace{(i(-), d')}_{\mathbb{y}_(d')i}\Big) \cong \Nat(d, \, d') \overset{\scriptsize\text{Yoneda lemma}}{\cong} \Nat\Big(\underbrace{(-, d)}_{\mathbb{y}_(d)}, \, \underbrace{(-, d')}_{\mathbb{y}_(d')}\Big)$$
The result follows by applying the dual version of the lemma with $i^{\rm op}$ ($i$ post-composed with the $\rm op$ functor) and $F ≝ \mathbb{y}_(d), \; G ≝ \mathbb{y}_(d')$:

Link to the svg picture
A: This is very very false.  You can simply take any natural transformation $Fi\to Gi$, and then define $\alpha_c$ for each $c\not\in A$ to be any map $Fc\to Gc$ at all.  It would be an enormous coincidence if the latter satisfied naturality.
For a simple example, let $C=D=Set$, $F=G=1_{Set}$, and let $A$ consist of just a singleton.  Any $\alpha$ is natural when restricted to $A$, but it's certainly not true that any collection of maps $X\to X$ for each set $X$ defines a natural transformation $1_{Set}\to 1_{Set}$
