Consider the set $X$, under the usual Euclidean metric, $d'(x, y) = |x - y|$. What do the open balls look like there? Given a centre $a > 0$, and radius $r$, we need to consider two cases: $r < a$ or $r \ge a$. In the former case, the ball takes the same shape is it would in $\Bbb{R}$:
$$(r - a, r + a).$$
In the latter case, the corresponding ball in $\Bbb{R}$ spills outside $X$ itself. We can't have $(r - a, r + a)$ as above, because such an interval would not be contained in $X$ (exception: if $a = r$, but this case can glom onto either case, and I'm trying to stay in keeping with the example you presented). The ball is the set of points $x \in X$ such that $d'(x, a) < r$, and it's not difficult to show that it comes to
$$(0, r + a),$$
which is (not so coincidentally) the Euclidean ball in $\Bbb{R}$, i.e. $(r - a, r + a)$, intersected with $X$, i.e. $(0, \infty)$. It's only when $r - a$ becomes negative that we need to worry about the intersection with $X$.
Now, the metric you are considering is $d(x, y) = |f(x) - f(y)|$, where $f : (0, \infty) \to (0, \infty)$ is defined by $x \mapsto \frac{1}{x}$. This map is injective, which makes $d$ a metric (the pullback metric). There's no surprises in how it works: it takes two points $x$ and $y$, maps them under $f$, and compares their distance in the image of $f$. Because $f$ inverts its argument, you can expect points very close to the origin to become very large, and vice-versa. Small distances near the origin will turn into large distances further away, and vice-versa.
The balls relate as you'd expect as well. If you want the ball $B$, centred at $a$, with radius $r$, with respect to $d$, you first find the ball $B'$, centred at $f(a)$, with radius $r$, with respect to $d'$. Then $B = f^{-1}(B')$. That is, $B$ contains all the points $x$ such that $f(x)$ lies within $r$ of $f(a)$... exactly as you would expect.
So, just as we have two cases under the Euclidean metric $d'$, we will have two cases under the metric $d$. If $r < f(a) = \frac{1}{a}$, then we have
$$B' = (f(a) - r, f(a) + r) = \left(\frac{1}{a} - r, \frac{1}{a} + r\right).$$
As such,
$$B = f^{-1}\left(\frac{1}{a} - r, \frac{1}{a} + r\right) = \left(\frac{1}{\frac{1}{a} + r}, \frac{1}{\frac{1}{a} - r}\right),$$
which simplifies to the form you've computed. When $r \ge \frac{1}{a}$, then
$$B' = (0, f(a) + r) = \left(0, \frac{1}{a} + r\right).$$
How do we compute $f^{-1}$ of this interval? We know that $x \in (0, \frac{1}{a} + r)$ if and only if $x < \frac{1}{a} + r$, which is true if and only if
$$\frac{1}{x} > \frac{1}{\frac{1}{a} + r} = \frac{a}{1 + ar}.$$
In other words, $x \in (\frac{a}{1 + ar}, \infty)$, i.e.
$$B = f^{-1}\left(0, \frac{1}{a} + r\right) = \left(\frac{a}{1 + ar}, \infty\right).$$
It should be no surprise that the points near $0$, which belong to $B'$, map under $f$ to the points "near" $\infty$.
Hope that helps.
