Suppose $F\colon\mathcal{D}\to\mathcal{C}$ is a functor, and $U\colon\operatorname{Cone}(F)\to\mathcal{C}$ is the functor sending a cone on $F$ to its vertex. I want to understand why
If $U$ has a colimit, then $F$ has a limit.
My idea so far is this. Suppose $(L,\alpha)$ is a colimit in $\mathcal{C}$ for $U$. Let $f\colon (M,r)\to (N,s)$ be a morphism of cones on $F$. For any $X\in\mathcal{D}$, $s_X\circ U(f)=r_X$, so $FX$ is a cocone of $U$ in $\mathcal{C}$. By the universal property, there exists a unique arrow $\beta_X\colon L\to FX$ such that $\beta_X\circ\alpha_M=r_X$ for all $X\in\mathcal{D}$.
My guess is $(L,\beta)$ is a limit for $F$. If $f\colon X\to Y$ is an arrow in $\mathcal{D}$, then $\beta_Y\circ\alpha_M=r_Y$, and $F(f)\circ\beta_X\circ\alpha_M=F(f)\circ r_X=r_Y$, so $\beta_Y=F(f)\circ\beta_X$. Hence $(L,\beta)$ is a cone over $F$.
Now if $(M,r)$ is any cone over $F$, then $\alpha_M\colon M\to L$ is a factorization through the cone $(L,\beta)$. But I don't see any way to conclude that $\alpha_M$ is unique. How can you conclude this last necessary piece?
You already proved that $(L, \beta)$ is a cone for $F$, so it is in $\operatorname{Cone}(F)$. That means that there is a coprojection $\alpha_L: L \to L$. By construction, each $\alpha_M: M \to L$ is a morphism of cones $\alpha_M: (M, r) \to (L, \beta)$, so $\alpha_L U(\alpha_M) = \alpha_M$ for all cones $(M, r)$. So we see that $\alpha_L$ is in fact a morphism of cocones, and must thus be the identity $Id_L$ because $(L, \alpha)$ is colimit.
Now let $f: (M, r) \to (L, \beta)$ be any morphism of cones. Then because $(L, \alpha)$ is a cocone for $U$ we have $\alpha_L U(f) = \alpha_M$, so by the discussion above $f = \alpha_M$, which proves uniqueness (which was indeed the only thing left to prove).