A good way to understand parametrisations is to try and plot one out from scratch. Trig functions can be a bit messy to start off with, so take something simpler like

and

. Pick different values of

and plot the corresponding points on an

-

graph. For example, for

you get

and

, so plot the point (1,2). Then imagine running through all values of

, and you'll see you end up with a continuous curve.
If you think of

as time, then this can be thought of as the motion of an object along a path. From experience, you can move along a path fast or slow, and that's why a parametrisation isn't unique. If

and

, you get the same curve in the

-

plane as in the example above, but this one is traced out 'faster', if you imagine

as time.
For your question, the given curve is drawn on

-

axes - it's a completely different curve to the hyperbola in question. The purpose of the

-

graph is to show what values

takes for given values of

. From the graph,

is positive for
)
. Again thinking of

as time, this tells you that in that period of time, the particle is travelling on the right side of the

-

plane, since the

-values are positive. The horizontal (

) coordinate is irrelevant in this regard, since left or right is determined by the value of

only.
But the particle could have just as easily travelled on the left side first. That's what you get if you let
)
. Then

is negative for
)
, but you still get the same path. The reason you get the same path is because the hyperbola is symmetric about the

-axis (as well as the

-axis). So by making all the

-values negative, you get the same path, but you get different points for the same values of

.