Well, to come back to the middle-school definition of an equation : an equation is an equality between two expressions which may contain unknowns (that is letters that represent quantities which we don't know yet). Solving an equation consist in finding all values by which we can replace the unknown and get a true equality. In this sense :
$$x + 3 = 8$$
is an equation (x is unknown) and its only solution is 5 because only $5+3$ equals $8$, not any other number plus 3.
Now that we have defined equations we may wonder how we can find their solutions in a mechanical way (and be sure that we found them all). To do this we know a certain number of way to find other equations that have the same solutions than the original (we say those equations are equivalent since they're true/false for the same values of the unknown), by choosing cautiously we may obtain an equation whose solutions are all obvious. Here by "subtracting 3" to both member of our first equation we get :
$$x + 3 = 8 \quad \Leftrightarrow \quad x + 3 - 3 = 8 - 3 \quad \Leftrightarrow \quad x = 5$$
And obviously, 5 is the only solution of this last equation... ($\Leftrightarrow$ means "is equivalent to" and often we make that implicit by putting equivalent equations on successive lines)
Now why do we get equivalent equations by subtracting 3 to each member ? Well that comes back to the fact that two equations are "equivalent" if they're true/false for the same values of the unknown : if you have two equal quantities and add/subtract the same amount to those two quantities, you'll still have equal quantities afterward, won't you ? And similarly if you start with different quantities, the results would still be different afterward.
That's why you can add or subtract the same number to both members of an equation when you want to solve it.
That's also why you can multiply or divide by the same non-zero number.
You can't multiply by zero because even if the initial equality was false it will becomes true : $10 \neq 5$ but $0 \times 10 = 0 \times 5$ since $0 = 0$.
There's a pretty simple criteria that dictate which operations you may use on both members simultaneously and obtains an equivalent equation : you can only use operations that you can reverse, that is there is an operation which allows you to come back to the initial value. "Adding 3" is reversed by "Subtracting 3", "Multiplying by 5" is reversed by "Dividing by 5" and so on.
$$7 + 3 = 10 \quad 10 - 3 = 7 \qquad -5 + 3 = -2 \quad -2 - 3 = -5 \qquad ...$$
$$3 \times 5 = 15 \quad 15 \div 5 = 3 \qquad -10 \times 5 = -50 \quad -50 \div 5 = -10 \qquad ...$$
Since there are a lot of operations with those characteristics, you can do a lot more than just add/subtract/multiply/divide by the same number but depending on your level you may not have seen any yet.
Note that there is a bit more to solving equations than I described here, in particular you may have cases where an equation is equivalent to several equations (and you'll need to be careful which logical operator you use in this case : and or or).
Your thinking is correct, but the book is wrong in general.
Assuming that $t\in [a,b]$ with $a<b$, then considering $\displaystyle \int_a^b y \frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y \dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $\dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y \dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $\dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y \dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $t\in [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $\displaystyle \int_a^b y \frac{dx}{dt}dt = \dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=r\cos\theta, y=r\sin\theta$. We find that $r\geq0$ and $0\leq \theta < \dfrac{\pi}{2}$. It's quite easy to show that $\dfrac{d\theta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=\cos\theta, y=\sin\theta$ for $\theta\in\left[0,\dfrac{\pi}{2}\right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-\dfrac{\pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y \dfrac{dx}{dt}$, nothing more.
Best Answer
Let's start with the simplest case, in which $h$ and $k$ are both $0$. Then the parabola is $$y^2 = 4ax$$ and the parametric equation is $$y=2at, \hskip{0.5in} x=at^2$$ and you can directly verify that $$y^2 = (2at)^2 = 4a^2t^2 = 4a(at^2) = 4ax$$ so everything works nicely.
Now let's consider the general case. We have $$(y-k)^2 = 4a(x-h)$$ If we introduce (temporarily) the new variables $u = x-h, v = y-k$ then the equation can be written as $$v^2 = 4au$$ which has exactly the form of the simple case we already considered. So we know the parametric equations for $u$ and $v$ are $$v=2at, \hskip{0.5in} u=at^2$$ But now we remember that $u = x-h$ and $v = y-k$. That means the parametric solutions are $$y-k=2at,\hskip{0.5in} x-h=at^2$$ which leads directly to the solution $$y=k+2at,\hskip{0.5in} x=h+at^2$$
Now let's look back and try to understand what's happening. The main idea is that the "shift" induced by replacing $x$ and $y$ with $x-h$ and $y-k$, respectively, corresponds to a shift in the solution that also replaces $x$ and $y$ with $x-h$ and $y-k$. The "plus" sign just comes from moving the negative terms to the other side of the equals sign.