Compute factor group $\dfrac{\mathbb{Z}_4 \times \mathbb{Z}_6}{\langle(2,3)\rangle}$ - Fraleigh p. 147 Example 15.11 (1.) Why's there a 'great temptation' to set $2 \bmod 4$ and $3 \bmod 6$ to 0? 
(2.) Why are you authorized to set $2 \bmod 4$ and $3 \bmod 6$ to 0?
$2 \bmod 4 \neq 0$ and $3 \bmod 6 \neq 0$, hence does this even make sense? 
(3.) How do you envisage and envision to work with $(1,0) + H = \{(1, 0), (3, 3)\}$?
I know the second paragraph starts with the Fundamental Theorem of Abelian Groups, but my course doesn't include this hence I can't use it. I also know:
$| \dfrac{\mathbb{Z}_4 \times \mathbb{Z}_6}{\langle(2,3)\rangle} | =  6 \times 4 = 24$, order $(\dfrac{\mathbb{Z}_4 \times \mathbb{Z}_6}{\langle(2,3)\rangle}) = lcm(4, 6) = 12.$ 

 A: This answers only your first question.  You should ask your three questions in three separate posts.
$\def\Z#1{\Bbb Z_{#1}}$
The “temptation” here arises from the way we understand quotient groups.  Consider the simpler example of $\Bbb Z/\langle 3\rangle $.  Intuitively, this is the result of taking the ordinary integers, $\Bbb Z$, and forcing $3=0$.  Of course $3\ne 0$, but we can ask what would happen if we were unable to distinguish between $3$ and $0$. The answer is that we get $\Z3$, which is a group in which $3$ is equal to $0$.  For example, in $\Z3$,  $2+1 = 0$ and $1+1+1=0$.  The construction of the quotient group as a group of cosets is a formalization of this intuitive idea.
For a different example, let $S_3$ be the set of symmetries of an equilateral triangle, $e$ be the identity symmetry, and $\rho$ be the reflection of the triangle along its vertical axis. $S_3/\langle\rho\rangle$ is the set of symmetries of the triangle if we choose to view the reflection as unimportant, or  less formally, if we pretend that $\rho = e$. The remaining symmetries are now just the rotations, and indeed $S_3/\langle\rho\rangle\cong \Z3$.
In $(\Z4\times\Z6)/\langle(2,3)\rangle$ we're doing  something similar: We start with $\Z4\times\Z6$ and ask what we would get if $(2,3) = (0,0)$.  It's easy to make the mistake that since $\Z4/\langle2\rangle \cong \Z2$ and $\Z6/\langle3\rangle \cong \Z3$ then  $(\Z4\times\Z6)/\langle(2,3)\rangle\cong \Z2\times\Z3$, but this is not the case, and this is what Fraleigh is warning you about.  $\langle(2,3)\rangle$ has order 2 in $\Z4\times\Z6$, so $(\Z4\times\Z6)/\langle(2,3)\rangle$ has 12 elements, but $\Z2\times\Z3$ has only 6, so it can't be right.
