Equations of motion for non-reciprocal spring-mass system

https://www.nature.com/articles/s41467-019-12599-3
The authors derive the equations governing a 1-D mass and spring system with non-reciprocal springs. The final equations, where $$u$$ is displacement,

$$\frac{1}{c}\frac{d^2u}{dt^2}-\frac{d^2u}{dx^2}+\frac{2\epsilon}{p}\frac{2du}{dx} = 0. \tag{1}$$

The equation is derived in the Methods section. The first steps are just a simple application of Newton's second law to a discrete system, and the following equation is obtained, where $$u_j$$ is displacement of mass $$j$$.

$$m\frac{d^2u_j}{dt^2} + k(1+\epsilon)(u_j-u_{j-1})+k(1-\epsilon)(u_j-u_{j+1})=0 .\tag{2}$$

Then the authors consider the continuum limit. $$u_j$$ becomes $$u(x)$$, which makes sense. For the displacement of the adjacent masses however, the equation below is presented,
$$u_{j\pm1} = u(x) \pm p\frac{du}{dx}+\frac{p^2}{2} \frac{d^2u}{dx^2} \tag{3}$$
($$p$$ is called the rest length)

1. What is the reasoning behind the transformation above? I get that something has to be substracted/added to $$u(x)$$, but I don't understand where the second and the third terms come from.

2. Also the authors state about equation (2),

In such a model, Newton’s action–reaction third law is broken, which means that in practice one needs to add local momentum at each site $$j$$ to realize such a system.

Why is the third law broken, and what is meant by "adding a local momentum"?

Your equation $$u_{j\pm1} = u(x) \pm p\frac{du}{dx}+\frac{p^2}{2} \frac{d^2u}{dx^2}$$ just looks like a second order Taylor polynomial. You are using this polynomial to approximate what $$u_{j\pm1}$$ is from $$u_j$$, where $$p=\Delta x$$. i.e. $$u_{j\pm1}=u(x\pm p)$$
Newton's Third Law is broken here because the system is non-reciprocal. i.e. the springs pull more in one direction than in the other direction. The force of mass $$j$$ on mass $$j+1$$ is not equal to the force mass $$j+1$$ exerts on mass $$j$$ (through the spring).