If I have understood it correctly, a factor group consists of all cosets of a subgroup $H$ of $G$. Since it is a requirement that $H$ is normal, left and right cosets are equivalent. I am asked to find the order of the element $5 + <4>$ in the factor group $\mathbb{Z}_{12}/<4>$. After some googling, I found the standard procedure to solve such tasks, and concluded with the following:
Since $<4> = \{0, 4, 8\}$ in $\mathbb{Z}_{12}$, and $5 +_8 5 +_8 5 + _8 5 = 8,$ we have that the order of $5 + <4>$ in $\mathbb{Z}_{12}$ is 4.
Note that at this point, I am shamefully reciting something I memorized while changing the numbers so that it'll fit the task. I have little to no understanding as to why this works. I do know that a factor group is a group of all cosets of the group $H$, in this case, $<4> \le \space \mathbb{Z}_{12}$, so if I want to explain my answer further, is my current intuitive guess good enough?:
We wish to find $$|<5 + \{0, 4, 8\}>|$$ in $\mathbb{Z}_{12}/<4>$. We have that $$<5 +\{0, 4, 8\}> \space=\{\{5,9,1\},\{10,2,6\},\{3,7,11\},\{8,0,4\}\}$$ Which is of order 4.
Your intuitive guess is pretty spot-on, and is much better than what Google told you to do.
Another way to tell that you have listed all of the cosets in the factor group is by noticing that the last element in the set (the one you got by adding the generator to itself 4 times) has 0 in it. That means it's the identity coset. Once you get the identity coset, you can stop writing down elements.
If you're having trouble picturing these cosets, you can also imagine the factor group construction as a type of "gluing". In the factor group, two elements are "glued" together if their difference is in the subgroup $H$. Then operations in the factor group are just operations in the original group, keeping the gluing in mind.
Now here's yet another way to tell that $5+<4>$ has order $4$: if you add it to itself $4$ times you get $8$ mod $12$. But $8$ is glued to $0$, since their difference is in $<4>$.