The major difference is in the choice of the Selberg Sieve weights. I strongly recommend reading [Maynard's paper][1]. It is well written, and the core ideas are nicely explained. For a longer exposition on Goldston Pintz and Yildirim's argument, Soundararajan has a nice [paper][2]. I also recommend reading this [expository paper][3] by Granville on the Maynard-Tao theorem, and the work of Zhang.

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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. 

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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 $$\sum_{x<n\leq 2x} \left(\sum_{i=1}^k \chi_{\mathcal{P}}(n)(n+h_i)\right)a(n)\geq \rho \sum_{x<n\leq 2x} a(n)$$ 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 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 left hand side, 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, 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 satisfying one of the required properties. Next, note that if each of $n+h_i$ is prime, since they are all greater than $x$, which is greater than $R$, and so the sum over $\lambda$ will consist only of $\lambda_1$. This means that $a(n)=1$ when each of the $n+h_i$ are prime, and so $a(n)$ 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 a difficult optimization takes place when choosing $\lambda_d$. Originally, the optimization in Selberg sieve involves diagonalizing a quadratic form, which leads 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. (As well as average over all sets of admissible $k$ tuples)

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. 


  [1]: http://arxiv.org/abs/1311.4600
  [2]: http://arxiv.org/abs/math/0605696
  [3]: http://www.dms.umontreal.ca/~andrew/CEBBrochureFinal.pdf