The major difference is the choice of the Selberg Sieve weights. I strongly recommend reading Maynard's paper. It is well written, and the core ideas are nicely explained.
In what follows, I'll give a brief explanation of what Selberg Sieve weights are, why they appear, and what was chosen differently in Maynard's paper compared to that of Goldston Pintz and Yildirim.
Let $\chi_{\mathcal{P}}(n)$ denote the indicator function for the primes. Given an admissible $k$-tuple $\mathcal{H}=\{h_1<h_2<\cdots<h_k\}$ the goal is to choose a non-negative function $a(n)$ so that
$$\sum_{x<n\leq 2x} \left(\sum_{i=1}^k \chi_{\mathcal{P}}(n)(n+h_i)\right)a(n)\geq \rho \sum_{x<n\leq 2x} a(n)\ \ \ \ \ \ \ \ \ \ (1)$$
for some constant $\rho>1$. If this is the case, then by the non-negativity of $a(n)$, we have that for some $n$ between $x$ and $2x$ $$\left(\sum_{i=1}^k\chi_{\mathcal{P}}(n)(n+h_i)\right)\geq \lceil\rho \rceil,$$ and so there are at least $\lceil\rho \rceil$ primes in an interval of length at most $h_k-h_1$. This would imply Zhangs theorem on bounded gaps between primes, and if we can take $\rho$ arbitrarily large, it would yield the Maynard-Tao theorem. However, choosing a function $a(n)$ for which equation $(1)$ this holds is very difficult. Of course we would like to take $$a(n)=\begin{cases}
1 & \text{when each of }n+h_{i}\text{ are prime}\\
0 & \text{otherwise}
\end{cases} ,$$
as this maximizes the ratio of the two sides, but then we cannot evaluate $\sum_{x<n\leq 2x} a(n)$ as that is equivalent to understanding the original problem. Goldston Pintz and Yildirim use what is known as Selberg sieve weights. They chose $a(n)$ so that it is $1$ when each of $n+h_i$ are prime, and non-negative elsewhere, in the following way:
Let $\lambda_1=1$ and $\lambda_d=0$ when $d$ is large, say $d>R$ where $R<x$ will be chosen to depend on $x$. Then set $$a(n)=\left(\sum_{d|(n+h_1)\cdots(n+h_k)} \lambda_d\right)^2.$$ By choosing $a(n)$ to be the square of the sum, it is automatically non-negative. Next, note that if each of $n+h_i$ is prime, then since they are all greater than $x$, which is greater than $R$, the sum over $\lambda_d$ will consist only of $\lambda_1$. This means that $a(n)=1$ when each of the $n+h_i$ are prime, and hence $a(n)$ satisfies both of the desired properties. Of course, we do not want $a(n)$ to be large when none, or even very few of the $n+h_i$ are prime, so we are left with the difficult optimization problem of choosing $\lambda_d$. The optimization in the classical application of the Selberg sieve to bound from above the number of prime-tuples involves diagonalizing a quadratic form, leading to
$$\lambda_d = \mu(d) \left(\frac{\log(R/d)}{\log R}\right)^k,$$
and so
$$a(n)=\left(\sum_{\begin{array}{c}
d|(n+h_{1})\cdots(n+h_{k})\\
d<R
\end{array}}\mu(d)\left(\frac{\log(R/d)}{\log R}\right)^{k}\right)^{2}.$$
This choice is not so surprising since the related function $\sum_{d|n}\mu(d) \left(\log(n/d)\right)^k$ is supported on the set of integers with $k$ prime factors or less. Goldston Pintz and Yildirim significantly modified this, and used a higher dimensional sieve to obtain their results.
The critical difference in the approach of Maynard and Tao is choosing weights of the form
$$a(n)=\left(\sum_{d_{i}|(n+h_{i})\ \forall i}\lambda_{d_{1}\cdots d_{k}}\right)^{2}.$$ This gives extra flexibility since each $\lambda_{d_1,\dots,d_k}$ is allowed depend on the divisors individually. Using this additional flexibility Maynard and Tao independently established equation $(1)$ for any $\rho>1$, with $k$ depending on $\rho$.
Here are four great resources where you can read more:
- Maynard's paper.
- Tao's blog posts.
- Granville's article on the Maynard-Tao theorem and the work of Zhang.
- Soundararajan's exposition of Goldston Pintz and Yildirim's argument.
Best Answer
1. The Proposition in the post is almost equivalent to the Conjecture, namely it implies that $$\frac{g_n}{\log g_n}\leq (2+o(1))\log n.$$ In particular, the Proposition (hence also the Conjecture) implies that $$g_n\ll\log n\,\log\log n.\tag{$\ast$}$$ This would contradict the common expectation that $g_n\gg (\log n)^2$ holds for infinitely many $n$'s.
2. On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to Ford-Green-Konyagin-Maynard-Tao (2014), and it states that $$g_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$ holds for infinitely many $n$'s.