An object of $S \downarrow U$ is an $S$-pointed graph, so there definitely is an initial object: which $S$-pointed graph has a unique $S$-embedding into every graph?
Answer: $S$ as a discrete graph with the natural $S$-pointing.
If you allow loops, there also is a terminal object: which $S$-pointed graph admits a unique $S$-homomorphism from any graph?
Answer: $S$ as a complete graph with all edges, including all loops.
I'm not exactly qualified to talk about size issues but, here goes.
Assuming both $(1)$ and $(2)$ make sense, they will describe isomorphic objects. To me the real, barebones definition is:
$(3)$ A limit of $F$ (regardless of how large $J,C$ are) is an object $L$ and a $J$-indexed family of arrows $(\phi_j:L\to F(j))$ satisfying the cone axiom such that any other cone $(\psi_j:K\to F(j))$ uniquely factors through the cone $\phi$.
Then $(1)$, $(2)$ and:
$(4)$ A limit of $F$ is a terminal object in a suitable category of cones
$(5)$ A limit of $F$ is an evaluation $R(F)$ where $R:C^J\to C$ is a right adjoint to $\Delta:C\to C^J$
$(6)$ A limit of $F$ is a representing object for the functor $\mathsf{Set}^J(\ast,C(-,F)):C\to\mathsf{Set}$
Are all variations on the same theme. We want (a) a way to restate $(3)$ as efficiently and cleanly as possible, because having clean organising language is an objective of category theory and (b) having different perspectives on the same thing, by viewing it as part of various different structures, is good because it broadens our mental horizons and allows us to apply theorems about said structures to our object of interest.
For example, I wouldn't ordinarily care about $(2)$ or its extremely close cousin $(6)$ but recently I encountered the concept of weighted limit. The perspective of $(6)$ allows us to immediately ask the question - what if we replace the trivial functor $\ast$ with a more interesting one?
A $W$-weighted limit of $F$ is a representing object for the functor $\mathsf{Set}^J(W,C(-,F))$.
That definition, and the resultant theory, wouldn't have been particularly possible if we had kept our eyes fixed on the barebones definition $(3)$. So I hope that's a sort of answer to the second part of your question.
As for size issues, $(3)$ is the essential idea; $(2)$ is a convenient way of encoding that idea, but it has the unfortunate drawback that in particular foundations it might not make sense for all categories. You're right it requires local smallness to have representability. If you're in a "large" situation you can still use $(3)$, which is morally the same but doesn't require you to construct a $\mathsf{Set}$-valued functor. Similarly the right adjoint might not always exist in general for $(5)$ and $(6)$ suffers from the same size issue... but all that does is make us state "limit" in the more boring manner of $(3)$, we don't really lose anything. The perspectives offered by the other definitions can still live in your head, even if verbatim they don't make sense. Note that even $(1)$ could be susceptible to size issues (though I don't know for sure) - does the comma category $(\Delta|F)$ exist? Is it small enough?
Possible reasons for restricting to limits in locally small categories is to (a) enjoy being able to literally state things like $(2),(6)$ and (b) because this "is usually true in practice". Also we usually only expect small limits to exist - that's implicit in the definition of complete category, for instance - because only this is true in our favourite prototypical category, $\mathsf{Set}$. But limits definitely can be, and should be, thought of in a size independent way. Just as long as you can talk about class functions and functors, you should be able to discuss $(3)$.
Best Answer
We can get a similar result to the proposition you mention, if we assume the diagram is connected and non-empty.
Proof. Let $F: I \to \mathcal{C}/C$ be some diagram. Denote by $U: \mathcal{C}/C \to \mathcal{C}$ the forgetful functor. Then as you already noted, we have a limiting cone $\lim UF$ in $\mathcal{C}$ with projections $p_i: \lim UF \to UF(i)$ for each object $i$ in $I$.
Now let $i$ be any object in $I$, then $F(i)$ is an object in $\mathcal{C}/C$, so it is some arrow $f_i: UF(i) \to C$ in $\mathcal{C}$. Define $\ell: \lim UF \to C$ as $\ell = f_i p_i$. This does not depend on the choice of $i$, which follows from the assumption that $I$ is connected. (This is the point where I hoped to draw a diagram, but I cannot make it work properly. So if someone else can, please do! In the meantime, try drawing it yourself on a piece of paper.) To see this, let $j$ be some object in $I$. There is a sequence of arrows between $UF(i)$ and $UF(j)$. For every step $k$ in this sequence we have a projection $p_k: \lim UF \to UF(k)$ and an arrow $f_k: UF(K) \to C$, such that everything commutes and $i$ and $j$ really give the same arrow $\ell$.
Now we do find a good candidate for the limit in $\mathcal{C}/C$, namely $\ell: \lim UF \to C$ together with the same set of projections $p_i$. This does indeed form a limit. Let $d: D \to C$ together with projections $q_i$ be some cone of $F$ in $\mathcal{C}/C$. Then $D$ together with $q_i$ forms a cone in $\mathcal{C}$. So there is an induced morphism of cones $u: D \to \lim UF$. Now we only need to check that $u$ is indeed an arrow in $\mathcal{C}/C$ as well. Let $f_i: UF(i) \to C$ be some object in the diagram of $F$, then because $q_i$ is an arrow in $\mathcal{C}/C$: $$ d = f_i q_i, $$ and since $u$ is a morphism of cones we have $q_i = p_i u$, so $$ f_i q_i = f_i p_i u, $$ finally by the definition that $\ell = f_i p_i$: $$ f_i p_i u = \ell u. $$ So summing up we have indeed $$ d = f_i q_i = f_i p_i u = \ell u, $$ as required. QED.
If the diagram is not connected, or if it is empty, we have no hope of the above proposition being true in general. Even if we assume $\mathcal{C}$ to have all limits. Consider the following two examples.
Example 1. No matter what category $\mathcal{C}$ and object $C$ we start with, the category $\mathcal{C}/C$ always has a terminal object and it is given by $Id_C: C \to C$. So if $\mathcal{C}$ already had a terminal object $1$, and we take $C$ to be non-terminal, then the forgetful functor does not preserve the terminal object.
Example 2. Let us consider $\mathbf{Set}$, the category of sets. Let us consider the set $\mathbb{N}$ of natural numbers, together with the subsets $E$ and $O$ of even and odd numbers respectively. We can naturally find $E$ and $O$ in $\mathbf{Set} / \mathbb{N}$ as well, by just considering the inclusions $E \hookrightarrow \mathbb{N}$ and $O \hookrightarrow \mathbb{N}$. The product of $E \times O$ in $\mathbf{Set}$ is just their cartesian product (with the obvious projections). The product in $\mathbf{Set} / \mathbb{N}$ does exist, but this is the empty set (with the empty function to $\mathbb{N}$)! This last part will be clear in a bit, when we prove that products in $\mathbf{Set} / \mathbb{N}$ are given by pullbacks in $\mathbf{Set}$ (so in this case, by the intersection $E \cap O$).
If we are just interested in $\mathcal{C}/C$ being complete, we have the following result.
This result does (implicitly) appear in most books about topos theory. When proving that for any topos $\mathcal{E}$ the slice topos $\mathcal{E}/X$, by some object $X$ from $\mathcal{E}$, is again a topos, one has to show that $\mathcal{E}/X$ is complete (although, technically this is about being finitely complete, but it easily generalises). This part of the proof only uses completeness of $\mathcal{E}$. For example, a proof can be found in Sheaves in Geometry and Logic by MacLane and Moerdijk, at the start of theorem IV.7.1. I will present a (sketch of a) proof here as well, so we can link it to the proposition at the start of this answer.
Proof. As mentioned in example 1 above, the category $\mathcal{C}/C$ always has a terminal object. By the proposition at the start of this answer, $\mathcal{C}/C$ has equalisers (and they are in fact 'the same' as in $\mathcal{C}$). So all we need to check is products. So let $(A_i \to C)_{i \in I}$ be a non-empty set of objects in $\mathcal{C}/C$. Form their wide pullback $P$ in $\mathcal{C}$. There is only one arrow $P \to C$ to be considered, and this will be the desired product in $\mathcal{C}/C$ (check this!). We have now shown that $\mathcal{C}/C$ has all small products and equalisers, so it is complete. QED.
We have essentially obtained a way to calculate limits in $\mathcal{C}/C$. For any diagram $F: D \to \mathcal{C}/C$ we obtain a diagram $F'$ in $\mathcal{C}$ by just 'forgetting' that we lived in $\mathcal{C}/C$. So I do not mean to just apply the forgetful functor here, because we want to keep all the arrows to $C$ in our diagram $F'$ (another way to describe this would be to apply the forgetful functor, and then add all the arrows to $C$ back in). Now we calculate the limit $\lim F'$ of $F'$ in $\mathcal{C}$. Since $C$ was in the diagram $F'$, we have a projection $\lim F' \to C$ and this will be the limit in $\mathcal{C}/C$.
The connection with the propisition at the start of this answer is that if $F$ is non-empty connected, we do not need to keep $C$ in the diagram to make things work.