Since we cannot factor out a $x-2$ term in the numerator for either of the $x \geq 0$ or $x < 0$ cases (the numerator is only a polynomial of degree $1$, after all), we can conclude that $x=2$ must be the sole vertical asymptote.
Now, as we discussed, the function $f(x)$ is defined to be piecewise rational as follows:
$$f(x) = \begin{cases} \frac{x+1}{x-2}, \quad x \geq 0 \\
\frac{-x+1}{x-2}, \quad x < 0 \\
\end{cases}$$
The definition of horizontal asymptote is as follows: we say that the function $f(x)$ has a horizontal asymptote $y = c$ if
$$\lim_{x \rightarrow +\infty} f(x) = c \quad \text{or} \quad \lim_{x \rightarrow -\infty} f(x) = c$$
The or part is what you should pay attention to, as we need only satisfy one of these conditions for $y = c$ to be a horizontal asymptote.
From this definition, we consider the piecewise definitions of $f(x)$ in each respective to domain to find the horizontal asymptotes. In your comment you said you can divide by the leading coefficient, but I think you meant to say that you divide by the leading variable (in this case, $x$) in both the numerator and denominator. This part is straightforward, and at once you can see that
$$y = 1 \quad \text{and} \quad y = -1$$
are indeed horizontal asymptotes as $x \rightarrow +\infty$ and $x \rightarrow -\infty$, respectively.