At the nLab page for limit, there is the mention of the $\mathrm{Lim}_D$ functor that sends every diagram to its limit.
What is the morphism function of $\mathrm{Lim}_D$ ?
I tried the following:
Given two diagrams $F$ and $G$, a morphism is a natural transformation, say $\sigma: F \rightarrow G$. As an image of $\sigma$ by $\mathrm{Lim}_D$, I would have looked after a morphism which is a member of $\sigma$ and links $\mathrm{Lim}_D(F)$ to $\mathrm{Lim}_D(G)$. However, as the limit of a diagram $F$ does not necessarily lies in the image of $F$, this construction seems wrong.
What is the morphism function of $\mathrm{Lim}_D$ ?
Let $C$ be a category having limits of shape $D$, being $D$ a category. Then we consider the category $[D,C]$ whose objects are the diagrams $D \to C$ and whose arrows are the natural transformations between them. We want to define a functor $\lim_D \colon [D,C]\to C$.
As you said, whenever $F$ is an object of $[D,C]$, it is the case that $\lim_DF$ is the limit of $D$ in $C$ (as the notation suggests) together with the cone $\alpha_F \colon \lim_DF \to F$ that exhibits $\lim_DF$ as a limit.
Now let us assume that $\beta$ is an arrow $F \to G$ of $[D,C]$. Then we have the limiting cones $\alpha_F\colon \lim_DF\to F$ and $\alpha_G \colon \lim_DG\to G$, right? Well, as we have $\beta \colon F \to G$, we also have a cone $\beta \circ \alpha_F\colon \lim_DF\to G$ over $G$. But since, among all the cones over $G$, $\alpha_G \colon \lim_DG\to G$ happens to be the teminal one (it is the limiting one!), there exists unique an arrow $\lim_DF \to \lim_DG$ making everything commute. We stipulate that $\lim_D\beta$ is precisely this arrow.
For instance, let us assume that $D=\{a,b\}$ without arrows. Then $\lim_D=\times$. So a diagram $F$ is a couple of objects $F_1,F_2$ of $C$, $\lim_DF=F_1\times F_2$ and $\alpha_F$ is just the couple ($\pi_1\colon F_1 \times F_2\to F_1$, $\pi_2\colon F_1\times F_2 \to F_2$), right?
Moreover, an arrow $\beta \colon F \to G$ is just a couple of arrows $(\beta_1\colon F_1\to G_1,\beta_2\colon F_2\to G_2)$. Hence, the couple $(\beta_1\circ\pi_1,\beta_2\circ\pi_2)$, that in the general case is $\beta\circ \alpha_F$, is a cone over $G=(G_1,G_2)$. But since the cone $(\pi_1 \colon G_1 \times G_2 \to G_1,\pi_2\colon G_1\times G_2 \to G_1)$ is the terminal one, there is unique an arrow $F_1\times F_2 \to G_1\times G_2$, the one that we usually indicate as $\beta_1\times \beta_2$, and that in the general case is $\lim_D\beta$.