Simple closed curve orientation definition 
The (simple closed curved) curve istraversed counterclockwise, and said to be positively oriented, if the region it (what is "it")? encloses is always to the left of an object as it (what is "it"?) moves along the path.

I don't understand this statement and found a counterexample.
What "object" are we talking about? I can think of a counterclockwise traversed curve where the region it enclosed is always right of an object as it moves along the path. I found this quote from a book and don't think the quote is correct. Just draw some arbitrary counterclockwise closed curve on your desk and remove every-single object to your right and we've found a counterexample because the set of objects that are to the right of the curve is empty so there's nothing to the left of the area enclosed.
 
 A: Imagine that an "object" (point) moves anticlockwise along the curve. Then the curve is positively oriented if the interior of the curve (the region enclosed by it) lies on the left hand side of the object.
Said in plainer terms, if you were to drive anticlockwise along a curved track, then the track is positively oriented if the region inside the curve is on your left.
Formalisation of the above
Say the simple closed curve $C$ in the plane encloses a point $O$. Let $P = \gamma(s)$ be a variable point which moves anticlockwise around $C$. The claim is that the point $O$ is on the left side of the tangent vector $T = \gamma'(s)$. In other words, that the angle between the vector $PO = \gamma(s) - O$ is acute; in other words that their cross product is positive. WLOG $O$ can actually be taken to be the origin, so we are only interested in the quantity $\gamma'(s) \times \gamma(s) = x'(s)y(s) - y'(s)x(s)$.
Moving anticlockwise means that the angle between $\gamma(s)$ and some fixed line (say the $x$ axis) is an increasing function of $s$. Apart from when $x=0$, this angle satisfies $\tan \theta(s) = y/x$, and therefore
$$\sec^2\theta(s) \theta'(s) = \frac{y'x-x'y}{x^2}$$
which shows that $O$ is on the left of $P$ if $\theta'(s) > 0$, in other words when $\gamma$ is positively oriented.
