If $(L, (p_D)_{D \in \mathcal{D}})$ is a limit of the functor $F : \mathcal{D} \to \mathcal{C}$, two morphisms $f, g : M \to L$ are equal as long as for every object $D \in \mathcal{D}$, $p_D \circ f = p_D \circ g$.
Proof. $f$ and $g$ are two factorizations of the cone $(M, (p_D \circ f)_{D \in \mathcal{D}})$ through the limit.
I am having trouble getting this proof. $(M, \dots)$ is a cone so for all $d : D \to D'$ in $\mathcal{D}$, $Fd \circ (p_D \circ f) = p_D \circ f$ and if $(M, (p_D \circ g)_{D \in \mathcal{D}})$ is another cone on $F$ then there is unqiue $u : M \to M$ such that $p_D \circ f \circ u = p_D \circ g = p_D \circ f$. I'm not seeing how to remove $p_D$ from the left of $f, g$.
For any cone, $(M,(q_D)_{D\in\mathcal{D}})$ there is a unique arrow $k : M\to L$ where $(L,(p_D)_{D\in\mathcal{D}})$ is the limiting cone such that $q_D=p_D\circ k$ for all $D\in\mathcal{D}$. This is the universal property of the limit. If $q_D=p_D\circ f$ for $f: M \to L$, then $k = f$. Of course, by exactly the same logic, if $q_D=p_D\circ g$ for $g:M\to L$ then $k = g$. Since $p_D\circ f = p_D\circ g$ for all $D\in\mathcal{D}$, $(M,(p_D\circ f)_{D\in\mathcal{D}})$ and $(M,(p_D\circ g)_{D\in\mathcal{D}})$ are the same cone and induce the same unique arrow $M \to L$.
I don't really care for talking about (co)cones though. We can view a cone, $(M,(q_D)_{D\in\mathcal{D}})$ over the diagram $F$ as a natural transformation $q:\Delta M\to F$. The universal property of the limiting cone says given a natural transformation $\Delta M\to F$ there exists a unique arrow $M\to L$, that is there is a bijection $\varphi:\mathsf{Nat}(\Delta M,F)\cong\mathsf{Hom}(M,L)$. The constraint that $q_D=p_D\circ k$ where $k\in\mathsf{Hom}(M,L)$ is handled by requiring this bijection to be natural in $M$. In this case, the cones $f$ and $g$ induce are exactly $\varphi^{-1}(f)$ and $\varphi^{-1}(g)$, and the equality assumption is just that $\varphi^{-1}(f)=\varphi^{-1}(g)$, from which it clearly follows that $f=g$ by applying $\varphi$ to both sides.