$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}[1]{\lfloor#1\rfloor}$The answer is yes to each of your two questions.

Let $a_n:=\phi(n)$. Then
\begin{equation*}
K(x)=\sum_{n\in\Z}a_n R(x-n).
\end{equation*}
Note that for all $j\in\Z$ we have $K(j)=a_j$ and $K$ linear (or, more exactly, affine) on the interval $[j,j+1]$. Also, $K$ is continuous. So, $K$ is obtained by the linear interpolation of the function $\Z\ni j\mapsto a_j$. In particular, $K$ is bounded.

So, for real $\ep>0$,
\begin{equation*}
\int_{1/\ep<|x|<\ep}K(x)\,dx=I_\ep+O(1/\ep),
\end{equation*}
where
\begin{equation*}
I_\ep:=\int_{|x|<\ep}K(x)\,dx
=\sum_{n\in\Z}a_n J_n,
\end{equation*}
\begin{equation*}
J_n:=\int_{-\ep}^\ep dx\,R(x-n).
\end{equation*}

Let now $N:=\lfloor\ep\rfloor$, so that $N\le\ep<N+1$. Then $J_n=\int_\R dx\,R(x-n)=1$ if $|n|\le N-1$ and $J_n=0$ if $|n|\ge N+2$. Also, $0\le J_n\le1$ for all $n\in\Z$. So,
\begin{equation*}
I_\ep
=\sum_{|n|\le N-1}a_n +O(|a_N|+|a_{-N}|+|a_{N+1}|+|a_{-N-1}|).
\end{equation*}
So, $I_\ep$ converges, since $N\to\infty$ (as $\ep\to\infty$), $\sum_{|n|\le N-1}a_n$ converges, and $|a_N|+|a_{-N}|+|a_{N+1}|+|a_{-N-1}|=O(1/N)\to0$. So, $\int_{1/\ep<|x|<\ep}K(x)\,dx$ converges.

This provides the positive answer to your first question.

The answer to your second question is also positive, that is, for some real $C_3$ and $C_4$,
\begin{equation*}
|K(x)|\le C_3|x|^{-1} \text{ if }x\ne 0 \tag{1}
\end{equation*}
and
\begin{equation*}
|K(x)-K(x-y)| \le C_4\frac{|y|}{|x|^2} \text{ if }|x|>2|y|>0. \tag{2}
\end{equation*}

Indeed, since $K$ is obtained by the linear interpolation of the function $\Z\ni j\mapsto a_j$ and the $a_j$'s are bounded, we see that the function $K$ is bounded and Lipschitz, so that without loss of generality $|x|>8$ in (1) and (2).

Now take indeed any real $x$ with $|x|>8$ and any real $y$ as in (2). Let
\begin{equation}
j:=\fl{|x|},\quad m:=\fl{|x-y|},
\end{equation}
so that $|j|\ge|m|\ge1+|x|/4$ and also $|j+1|\ge|x|/4$.

So, by the linear interpolation observation and the condition $|a_n|\le C_1/|n|$ for $n\ne0$,
\begin{equation*}
|K(x)|\le|a_j|+|a_{j+1}|\le C_1(|j|^{-1}+|j+1|^{-1})\le8C_1|x|^{-1},
\end{equation*}
which verifies (1).

Next, by the linear interpolation observation and the condition $|a_{n+1}-a_n|\le C_2/n^2$ for $n\ne0$, the function $K$ is Lipschitz on $[m,\infty)$ and on $(-\infty,-m]$ with Lipschitz constant $C_2/(m-1)^2\le4C_2/x^2$.
Also, $|x|\ge j\ge m$, $|x-y|\ge m$, and, by the condition $|x|>2|y|>0$ in (2), $x$ and $x-y$ are of the same sign. So, either both $x$ and $x-y$ are in $[m,\infty)$ or they are both in $(-\infty,-m]$.

Therefore and because the function $K$ is Lipschitz on $[m,\infty)$ and on $(-\infty,-m]$ with Lipschitz constant $4C_2/x^2$, (2) follows.