Sums of independent variables where f_X is a function of y? 
I don't understand example 3a or the theory above before it. How can an independent variable have the constraint that $f_X$ is a function of y as introduced in the theory that leads up to (3.1)? From the def to the right above it seems that $f_X$ can not be a function of y and $f_Y$ can not be a function of x if they are independent. Asa further illustration I have added an example 2f where they look at the function 24xy defined on $0<x<1 \hspace{0.5cm}   0<y<1 \hspace{0.5cm}   0<x+y<1$ and find it to be dependent. In the green box below I have marked in purple the part that I believed defined 24xy to be dependent. Why is 24xy dependent while the theory that leads to (3.1) describes independent relations?
And why do they differentiate with $\frac{d}{da}$ to obtain $f_{X+Y}$? 
The theory is taken from Ross, A first course in probability.
 A: In equation (3.1), $X$ and $Y$ are independent and so the joint distribution for the two random variables is the product $f(x,y)=f_X(x)f_Y(y)$. The pdf $f_X(x)$ depends only on $x$. An integration is performed, over the region where $x+y<a$. The limits of integration for $x$ therefore depend on $y$ and $a$. Remember that in general, the cdf $F(b)$ is defined by 
$$F(b)=P(X<b)=\int_{-\infty}^b f(x)\,dx$$
whereby $b$ is simply specifying the interval under consideration; $b$ is not a defining parameter of the r.v. $X$. In this case $b=a-y$; this does not imply that the $X$ itself depends on $a$ or $y$.
In terms of your second question, by definition,
$$f(b)={d\over db}F(b)$$
so the derivative is exactly what is needed to find the pdf if you know the cdf.
A: In answer to your updated question, remember that for independent $X$ and $Y$, the joint pdf
$$f(x,y)=f_X(x)f_Y(y)$$
and this is not the case for the second part of 2f because the definition of the joint pdf is not only that $f(x,y)=24xy$ but also that $f(x,y)=0$ if $x+y$ is outside the range $[0,1]$. It is not possible therefore to express the joint pdf as a product of $f_X(x)$ and $f_Y(y)$ such that $f_X(x)$ depends only on $x$. If this is not clear, consider some examples: if $y=0.9$ then the pdf for $X$ alone (the marginal pdf) is zero except in the range $(0,0.1)$ whereas if $y=0.5$, the pdf for $X$ alone is non-zero in the range $(0,0.5)$. It should be clear that the distribution for $X$ depends on $y$, so the two random variables are dependent.
