How do you derive a normal vector from the equation of a line? I've seen that to find a normal vector to a line such as $3x+4y-1=0$ people take the coefficients and say $(3,4)$ is a normal vector?

Why does this work? How are the coefficients related to the normal?
Two vectors $a,b$ are normal iff $a\cdot b=a_1b_1+...+a_2b_2=0$.

If you have a vector, $(x,y)$ and you want to find vectors that are normal to it you want to find a vector $(a,b)$ such that $ax+by=0$. So a normal vector to the line $3x+4y=0$ is simply $(3,4)$. 

If you have a constant on the right side, it just moves the line up or down. It doesn't change anything else. So a normal vector will still be $(3,4)$.
You want to understand this stuff? There's an infinite number of lines (in 2 dims) that rise 7 whenever they run 3. Draw any one of them. Call it L1.

Now, there's an infinite number of lines that are perpendicular to L1. Draw any one. Now use geometry (basic triangle laws) to answer this about this second line: how much does it rise when it runs 7?

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Added: It's not clear what level you are at, but you will need proficiency with both the $y = mx + b$ and $Ax + By + C = 0$ forms of equations of a line (and converting between them) in order to benefit from my suggestion.