Optimizing coin splitting - Is this algorithm as fast as I think?
You have made significant progress on this problem. Your final conclusion, "the overall algorithm asymptotically requires $\Omega(n^2)$ steps" is likely to be correct as well.

Analysis of Your Argumentation

Here are some unclear or unconventional or inconsistent notations found in your post, which makes it difficult to understand or justify your arguments about time-complexity.


Does "$n|2$" mean $n$ is even? The standard convention for for "$a$ divides $b$" is $a|b$ according to math stackexchange answers and my years of practicing number theory. So $2|n$ is the prefered notation for $n$ is even. Anyway, in current case, I strongly prefer the plain English, $n$ is even.
"use tuples to denote ranges of values. $(i,j)$ specifies the value range $v_i,\cdots,v_j$.". People use $(i,j)$ generally to mean the pair of elements $i$ and $j$ or an open interval with ends $i$ and $j$. I would suggest you use $v_{i..j}$, where the two-dot notation $i..j$ are the convention for an integer interval. 
"create a DAG $G = (V, E, \gamma)$". Do you mean an edge-weighted DAG?
"Create a vertex $v$ named $(1,n)$", which does not sound like English. It is better to write "Create a vertex $v$ that represents $1..n$".
"Create or refer to the previously created vertices $v_1,v_2,v_3$, named $(2,n−1),(3,n),(1,n−2)$ respectively". Although I can probably guess your intention, I am puzzled by the notations. Did you not notice that $v_1, v_2, \cdots, v_n$ appears as the coin values in the quoted problem? It looks like you want to use "$(2,n-1)$" as a name, which would be very confusing, since I have never seen any name of a person, a item, a variable or whatever starts and ends with a parenthesis for years except for very few social online accounts.
"Create the edges $e_1=(v,v_1),\gamma(e_1)=v_1$ and $e_2=(v,v_2), \gamma(e_2)=\max\{v_1,v_n\}$ and $e3=(v,v_3),\gamma(e_3)=v_n$". Now I am more into the wanderland. Is the codomain of $\gamma$ the vertices or the indices or the coin values?
"Iterate through the vertices $v_{\lbrace 1,2,3\rbrace}$". It is better to just write $v_1, v_2, v_3$.
"the sequence of coins denoted by the $v$". Didn't you say $v$ is a vertex?


Now Let us check your computation of the time-complexity.


"The number of edges in the created graph is bounded as follows: $|E|\le3|V|$. The number of vertices is denoted by the following sum:  $1 + \sum_{i=0}^{\frac{n}{2}} 2i + 1 \leq n^2$". It looks like your estimates are correct, even though they can be tighter.
"We can determine the maximum score for player 1 by 'backpropagating' the edge weights through the graph, which takes $| V | * | E |$ steps. This makes the overall algorithm asymptotically require $\Theta(n^2)$ steps". Since it is not clear how your backpropagating works exactly, it is hard to justify the number of steps. However, since $| V |* | E | \le |V| * (3 |V|) \le 3 n^4 $, you can only say that your algorithm takes $O(n^4)$ time.


In summary, even assuming that your algorithm is indeed a correct way, there is not enough clear information to fully justify or disprove your time-complexity argumentation. There are quite some room for you to improve the description of your algorithm and the time-complexity argumentation.

A Reference Solution

For simplicity, I will assume $n$ is even. Let $f(i,j)$ be the maximal score reachable by the first player if the two players start the game with the coins sequence $v_i, v_{i+1}, \cdots, v_j$. 

Then we have the base cases,
$$f(i, i+1)=\max(v_i, v_{i+1})$$
and recurrence relation
$$ f(i,j) =\max(v_i+\min(f(i+2,j),\ f(i+1,j-1)),\ v_j +
\min(f(i+1,j-1),\ f(i,j-2)))$$

The parameter pair $(i,j)$ of any $f(i,j)$ that appears in the course of computation of $f(1,n)$ will be such that $1\le i\le j\le n$ and $(i-1)+(n-j)$ must be even. The number of all such parameter pairs are $n^2/4$. The base case and the recurrence relation together shows that we need some $O(1)$ operations to obtain $f(i,j)$ for a new parameter pair $(i,j)$, assuming that the usual memorization technique in dynamic programming is used. So the time-complexity of computing $f(1,n)$ is $\Theta(n^2/4)=\Theta(n^2)$.