Consider the formula for the linearization of a function $f(\mathbf{x})$ about a point $\mathbf{x}_{e}$, where $\mathbf{x}, \mathbf{x}_{e} \in \mathbb{R}^{n}$:
$$f(\mathbf{x}) \approx f(\mathbf{x}_{e}) + \nabla f|_{\mathbf{x}_{e}} \cdot (\mathbf{x}-\mathbf{x}_{e})$$
I will only write out the math for problem 1, but problem 2 is hardly any different. So, we have:
$$\ddot{x} = f(x, \dot{x})$$
$$\ddot{x} = -\lambda \frac{|\dot{x}|\dot{x}}{2} - U \lambda (x-x_{sp})$$
Thus, when considering the linearization about the point $\mathbf{x}_{e} = [x_e \ \dot{x_e}]^T = [x_{sp}\ 0]^T$ , we have:
$$f(\mathbf{x}_e) = 0$$
$$\nabla f|_{\mathbf{x}_e} \cdot (\mathbf{x} - \mathbf{x}_e) = \frac{\partial f}{\partial x}|_{x_{sp}} \cdot (x-x_{sp}) + \frac{\partial f}{\partial \dot{x}}|_{0} \cdot (\dot{x} - 0) $$
We can see the first term $\frac{\partial f}{\partial x}|_{x_{sp}} \cdot (x-x_{sp})$ is simply $-U\lambda (x-x_{sp})$. As the second term is the point of interest in this question, we will work it out in a more verbose manner:
$$\frac{\partial f}{\partial \dot{x}} = -\lambda \frac{2sgn(\dot{x})\dot{x}}{2} = - \lambda|\dot{x}|$$
$$\frac{\partial f}{\partial \dot{x}}|_{0} = -\lambda |0| = 0$$
$$\frac{\partial f}{\partial \dot{x}}|_{0} \cdot (\dot{x} - 0) = 0$$
Thus, our linearized differential equation in the neighborhood of $\mathbf{x}_{e}$ becomes:
$$\ddot{x} + U\lambda(x-x_{sp}) \approx 0 $$
The intuition as to why we see certain terms disappear is that as we approach neighborhoods of the origin, terms with degree $\gt 1$ (i.e. $|\dot{x}|\dot{x}$), or more generally first order Taylor expansions that are still functions of the original variable (for the most part), have derivatives which are negligible in a neighborhood of the origin - unlike a linear term. Remember that $f(\mathbf{x}_{e}) = 0$, so we should not see any $i^{th}$ order term where $\lim\limits_{\mathbf{x} \to 0} f^{(i)}(\mathbf{x}) = \infty$, so this removes the possibility of a first-order term like $1/\mathbf{x}$.
Finally, consider a point $\epsilon \approx 0$. Given functions $g(\epsilon) = \epsilon$ and $h(\epsilon) = \epsilon^{2}$, we note $g'(\epsilon) = 1$ and $h'(\epsilon) = 2\epsilon$. Since we know $\epsilon \approx 0$, $h'(\epsilon) \approx 0$ as well. Thus if we have the equation $f(x) = g(x) + h(x)$ modeling our dynamics, we would see that the linear term "dominates" close to the origin.
Edit: Let $y_{r} = -\frac{1}{U\lambda}(2U + \frac{\lambda r_{0}^{2}}{2})$ and $\dot{y}_{r} = 0$
Consider (12) rewritten as:
$$\ddot{y} = f(y, \dot{y})$$
$$\ddot{y} = -\frac{\lambda}{2}(\dot{y}+2r_{0})\dot{y} - U\lambda y - 2U - \frac{\lambda r_{0}^{2}}{2}$$
$$\ddot{y} = -\frac{\lambda}{2}\dot{y}^{2} - \lambda r_{0}\dot{y} - U\lambda y - 2U - \frac{\lambda r_{0}^{2}}{2}$$
Let us now consider $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial \dot{y}}$:
$$\frac{\partial f}{\partial y} = -U\lambda$$
$$\frac{\partial f}{\partial \dot{y}} = -\lambda\dot{y} - \lambda r_{0}$$
Now, let us consider the equilibrium point $\mathbf{y}_e = [y_{r} \ \dot{y}_{r}]^{T}$. So, using the linearization formula from part one, we get:
$$f(y_{r}, \dot{y}_{r}) = 0$$
$$\frac{\partial f}{\partial y}|_{y_{r}} \cdot (y-y_{r}) = -U\lambda (y-y_{r}) = -U\lambda y - 2U - \frac{\lambda r_{0}^{2}}{2}$$
$$\frac{\partial f}{\partial \dot{y}}|_{\dot{y}_{r}} \cdot (\dot{y}-\dot{y}_{r}) = -(\lambda \dot{y}_{r} - \lambda r_{0})\cdot (\dot{y}-\dot{y}_{r}) = -\lambda r_{0} (\dot{y}-0) = -\lambda r_{0}\dot{y}$$
Now, I would assume the author drops the constant terms as they have no impact on the stability characteristics at the point $\mathbf{y}_e$ (they only determine where the equilibrium point will be, not what the convergence actually looks like). After dropping these terms, we are left with:
$$\ddot{y} + \lambda r_{0}\dot{y} + U \lambda y \approx 0$$
in a neighborhood of the point $\mathbf{y}_e$.
Resources:
Question2: how the equatoin 12 is linearized in the neighborhood of the equilibrium point $y = y_r$ ? what operations on $y \ \dot y \ {\dot y}^2 \ \ddot y$ and the constant ?
