Before giving a lenghty introduction, I'd like to actually just ask one thing.
We are given the object part Ev$_0$ of the evaluation functor $\mathcal{C} × [\mathcal{C}, \mathcal{D}] → \mathcal{D}$. I'm trying to define the morphism part of Ev: $\text{ Ev}_1$. It should map a pair
$$(f : C → C',~~ \mu : (F : \mathcal{C} → \mathcal{D}) \Rightarrow (F': \mathcal{C} → \mathcal{D}))$$ to a morphism of $\mathcal{D}$. I have an idea (namely, the morphism $$F'(C) \circ \mu_C : F(C) \to F'(C').)$$
Now, all this seems to be well-defined by naturality, but now I have to show $\text{Ev}_1$ preserves composition (of pairs of an object of $\mathcal{C}$ and a natural transformation $\mu$. That is, $$\text{Ev}_1((g : D → D', \nu : G \Rightarrow G') \circ (f: C → C', \mu : F \Rightarrow F')) = \text{Ev}_1((g, \nu ) \circ \text{Ev}_1((f, \mu)).$$
Now I know there are two way to compose natural transformations: horizontal and vertical. But horizontal composition seems not to apply because we are only working with two categories, $\mathcal{C}$ and $\mathcal{D}$. That leaves vertical composition, which is a but more arduous, but I've gone through it before.
Now comes the real question: in general, for any functor $F : \mathcal{C} → \mathcal{D}$, does the equality $F(g \circ f) = F(g) \circ F(f)$ only have to be checked for composable morphisms $f, g$? Or must we make them composable..? (I'm not sure we can always do this, certainly not in certain examples.)
Either way, suppose I'm given any two morphisms $f : A₁ → A₂$ and $g : B₁ → B₂$ in $\mathcal{C}$, and any two natural transformations $\mu : F \Rightarrow F'$ and $\nu : G \Rightarrow G'$ in $[\mathcal{C}, \mathcal{D}]$, is it correct to define
$$ (g : B₁ → B₂, \nu : G \Rightarrow G') \circ (f : A₁ → A₂, \mu : F \Rightarrow F') := (G'F'(f) : F(C) → G'F'(C'), \nu \circ \mu : G \circ F \Rightarrow G' \circ F')?$$
EDIT
In concatenating the naturality squares for $\mu$ and $\nu$, I found out that, having chosen the composable $\mathcal{C}$-morphisms $f : C → C'$ and $g : C' → C''$, I should probably have taken $h : = g \circ f : C → C''$ to show that the concatenated naturality squares combine to indeed form another naturality square, but with vertical arrows $F(h) : F(C) → F(C'')$, $F'(h) : F'(C) → F'(C'')$, and $F''(h) : F''(C) → F''(C'')$, correct?
Start with two composable morphisms in $\mathcal C \times [\mathcal C,\mathcal D]$: $$ (A,F) \stackrel{(f,\,\mu)}\longrightarrow (B,G) \stackrel{(g,\, \nu)}\longrightarrow (C,H). $$
Then, as you said, $\text{Ev}_1((f,\mu))$ is the diagonal morphism of the first diagram below, and $\text{Ev}_1((g,\nu))$ is the diagonal morphism of the second one. $$ \require{AMScd} \begin{CD} F(A) @>{F(f)}>> F(B) \\ @V{\mu_A}VV @VV{\mu_B}V \\ G(A) @>>{G(f)}> G(B) \end{CD} \qquad \quad \require{AMScd} \begin{CD} G(B) @>{G(g)}>> G(C) \\ @V{\nu_B}VV @VV{\nu_C}V \\ H(B) @>>{H(g)}> H(C) \end{CD} $$
On the other hand, the composition $(g,\nu)(f,\mu)$ of $(f,\mu)$ with $(g,\nu)$ is $$ (A,F) \stackrel{(gf,\,\nu\mu)}\longrightarrow (C,H), $$ and $\text{Ev}_1((gf,\nu\mu))$ is the diagonal morphism of the following diagram. $$ \require{AMScd} \begin{CD} F(A) @>{F(gf)}>> F(C) \\ @V{(\nu\mu)_A}VV @VV{(\nu\mu)_C}V \\ H(A) @>>{H(gf)}> H(C) \end{CD} $$ But we can split the latter diagram as $$ \require{AMScd} \begin{CD} F(A) @>{F(f)}>> F(B) @>{F(g)}>> F(C) \\ @V{\mu_A}VV @V{\mu_B}VV @V{\mu_C}VV \\[-1mm] G(A) @>{G(f)}>> G(B) @>{G(g)}>> G(C) \\ @V{\nu_A}VV @V{\nu_B}VV @V{\nu_C}VV \\[-1mm] H(A) @>{H(f)}>> H(B) @>{H(g)}>> H(C) \end{CD} $$ which means $$ \text{Ev}_1((g,\nu)(f,\mu)) = \text{Ev}_1((g,\nu)) \text{Ev}_1((f,\mu)). $$