Before answering the 2 questions I would like to provide some background.
Notes
As the Lagrangian is the function $L(x_i,\dot{x}_i)$, which is the function of 2 variables $x_i,\dot{x}_i$. Thus the partial derivative expansion is
\begin{equation}\tag{1}\label{eqn:1}
\delta L = \sum_i\frac{\partial L}{\partial x_i}\delta x_i + \sum_i\frac{\partial L}{\partial\dot{x}_i}\delta\dot{x}_i
\end{equation}
where $\delta$ represents infinitestimal change. This reads as, "the change in $L$ caused by the infinitestimal change in position: $\delta x_i$ and infinitestimal change in velocity: $\delta\dot{x}_i$".
Now at first glance we think that $\delta\dot{x}_i$ is the time derivative of $\delta x_i$, but these displacements must be seen as independent changes. This is confusing because of the notation, but $x_i$ and $\dot{x}_i$ are explicit variables in $L$ and because of partial derivatives, $L$ doesn't know how the connection between $x_i$ and $\dot{x}_i$ and treats them as independent.
Example, let
\begin{equation}
L = x +3\dot{x}\\
\delta L = 1\cdot\delta x + 3\cdot\delta\dot{x}
\end{equation}
renaming the variables shows that it is treated simply as variables:
\begin{equation}
L = x +3y\\
\delta L = 1\cdot\delta x + 3\cdot\delta y
\end{equation}
It is only when we introduce
\begin{equation}\tag{2}\label{eqn:2}
\delta\dot{x}_i = \delta\frac{dx_i}{dt} = \frac{d}{dt}\delta x_i
\end{equation}
is $\delta\dot{x}_i$ the time derivative of $\delta x_i$.
Now for Eq: ($\ref{eqn:1}$) we can't independently change $\delta\dot{x}_i$ because we know that $\delta\dot{x}_i$ the time derivative of $\delta x_i$, and by changing one of these variables will change the other to maintain the relation Eq:($\ref{eqn:2}$)
So now $\delta L$ actually has a condition, so we can rename it to keep track of it: $\delta L_{\varepsilon}$, simily $\dot{x}_{i\varepsilon}$ for $\dot{x}_{i}$ hence:
\begin{equation}\tag{3}\label{eqn:3}
\delta L_{\varepsilon} = \sum_i\frac{\partial L}{\partial x_i}\delta x_{i} + \sum_i\frac{\partial L}{\partial\dot{x}_i}\delta\dot{x}_{i\varepsilon}
\end{equation}
Now in the question, they use the fact that under translation of the entire system, $\delta L = 0$. So considering only varied displacement, we would initially get
\begin{equation}\tag{4}\label{eqn:4}
\begin{aligned}
\delta L &= \sum_i\frac{\partial L}{\partial x_i}\delta x_i + \sum_i\frac{\partial L}{\partial\dot{x}_i}\cdot 0\\
&= \sum_i\frac{\partial L}{\partial x_i}\delta x_i\\
&= 0
\end{aligned}
\end{equation}
where $\delta\dot{x}_i = 0$, as we not varing this.
But because we want the $\delta$ with the condition, thus
\begin{equation}\tag{5}\label{eqn:5}
\delta L_{\varepsilon} = \sum_i\frac{\partial L}{\partial x_i}\delta x_{i} + \sum_i\frac{\partial L}{\partial\dot{x}_i}\delta\dot{x}_{i\varepsilon} = 0
\end{equation}
and because $\delta x_{i}$ is not explicitly or implicitly a function of time(we are not doing variation caculus here, so $\delta x$ is not defined as $\delta x = \frac{\partial x}{\partial \alpha}d\alpha = \eta(t)d\alpha$ like in chapter 6),
\begin{equation}\tag{6}\label{eqn:6}
\delta\dot{x}_{i\varepsilon} = \delta\frac{dx_{i}}{dt} = \frac{d}{dt}\delta x_{i} = 0
\end{equation}
hence
\begin{equation}\tag{7}\label{eqn:7}
\begin{aligned}
\delta L_{\varepsilon} &= \sum_i\frac{\partial L}{\partial x_i}\delta x_{i} = 0\\
&= \delta L
\end{aligned}
\end{equation}
which turns outthe same if considering only varied displacement.
Answer to question 1
They did actually repeat the same logic here. for Eq:(7.140) they were working with an infinitestimal change in angle $\delta \theta$, where
\begin{equation}\tag{8}\label{eqn:8}
\delta \textbf{r} = \delta \boldsymbol{\theta} \times \textbf{r}
\end{equation}
and the time derivative is:
\begin{equation}
\begin{aligned}
\frac{d}{dt}\left(\delta \textbf{r}\right) &= \frac{d}{dt}\left(\delta \boldsymbol{\theta}\right) \times \textbf{r} + \delta \boldsymbol{\theta} \times \frac{d}{dt}\left(\textbf{r}\right)\\
&= 0\times \textbf{r} + \delta \boldsymbol{\theta} \times\dot{\textbf{r}}\\
&= \delta \boldsymbol{\theta} \times\dot{\textbf{r}}
\end{aligned}
\end{equation}
Thus this is the same logic applied for Eq:(7.132) where $\delta \theta$ is not explicitly or implicitly a function of time.
Answer to question 2
First we can define $x(t,\alpha) = x(t) + \alpha\eta(t)$.
A displacement in the system leads to $x(t,\alpha) = x(t) + \alpha\eta(t) + \delta x$. Thus the variation $\eta(t)$ is still unaffected and equate to $0$ at the end points.
For $\delta L$, depending on how it modeled, it may or may not change during the $\delta x$ shift
