vertical asymptotes appear when the function tends towards plus or minus infinity,
such as when a 0 would appear in a denominator somewhere.
In your transformed function $f(x)$ appears in a denominator. Are there any places where $f(x)=0$ ?
To find horizontal asymptotes you need to know the behavior of the function as $x \to \pm \infty$.
The graph given doesn't display this so you aren't able to determine any horizontal asymptotes from the information given.
O.K. let's look at the actual function:
$f(x)=\begin{cases}1 &: x\in [-4,-2] \\ -x-1 &: x\in [-1,0]\\2x-1 &: x\in[0,1]\end{cases}$ so let $g(x)=\dfrac{1}{f(x)}+1\begin{cases}2 &: x\in[-4,-2] \\ \dfrac{x}{x+1} &: x\in [-2,-1)\cup (-1,0]\\\dfrac{2x}{2x-1} &: x\in [0,0.5)\cup (0.5,1]\end{cases}$
Now you should see what you need.
You can take the approach Plato took and list everything out explicitly but I think that misses the point of the problem.
The asymptotes, now that $f(x)$ is in the denominator after the transform, will be where $f(x)=0$
These points are easily seen on the graph as $x=-1,~x=\dfrac 1 2$
and indeed if you look at the graph of $\dfrac {1}{f(x)}+1$
you see that this is indeed where the vertical asymptotes are.