What does the derivative mean in least squares curve fitting? Background:
When trying to fit a straight line to a bunch of data points, if we use y=ax+b, then the error e at every data point will be e = Y - y, where Y is the data point's position on the y axis.  
The next step is to take the sum of the squares of the error: S = e1^2 + e2^2 etc.  
Then we substitute as S = summation((Yi - yi)^2) = summation((Yi - (axi + b))^2).  
To minimize the error, we take the derivative with the coefficients a and b and equate it to zero.  
dS/da = 0 and dS/db = 0.  
Question:
I don't understand why we take a derivative of the coefficient. In y=mx+c, the coefficient is m, and that's the slope. We would take the derivative of the slope to find the rate of change of slope at any instant (although more correctly, it'd be rate of change of y wrt x at any instant).  


*

*But for curve fitting, what does it mean to say dS/da and dS/db? What
are we actually measuring when we say rate of change of a summation
of error with respect to a or b?

*Why does a coefficient like a or b become so important? Why and how is a coefficient so prominently related to error? How does it affect anything?

*I understand dS/da is equated to zero to mean that we want the sum of
errors to tend to zero, but how does equating it actually make it
tend to zero?

 A: Least squares
You want to find the parameters for a model which best describes the data. Furthermore, you have specified that you want the best fit with respect to the $l_{2}$ norm. Let's look at a simpler case which allows us to explore the consequences of these choices.

Find the average
Computing the average is computing a least squares solution. Mathematical details follow.
Input data
Start with a sequence of $m$ measurements $\left\{ x_{k} \right\}^{m}_{k=1}$. Perhaps these numbers are test scores for a class.
Model
How would you characterize the performance of the class? Your model is simple:
$$
 y(x) = \mu
$$
We know this number $\mu$ will be the average.  The free parameter in the least squares fit is the constant $\mu$.
Least squares problem
The least squares problem minimizes the sum of the squares of the differences between the measurement and the prediction. Formally,
$$
 \mu_{LS} = \left\{ 
 \mu \in \mathbb{R} \colon 
 \sum_{k=1}^{m} \left( x_{k} - \mu \right)^{2} \text{ is minimized}
 \right\}
$$
The function
$$
\sum_{k=1}^{m} \left( x_{k} - \mu \right)^{2}
$$
is called a merit function. This is the target of minimization. 
Least squares solution
We know how to find extrema for functions: we look for the points where the derivatives are $0$. Remember, the parameter of variation here is $\mu$.
$$
  \frac{d}{d\mu} \sum_{k=1}^{m} \left( x_{k} - \mu \right)^{2} = 0
\tag{1}
$$
Sticklers may protest that this finds extrema, yet we need minima. These fears will be allayed by posting the question "How do we know that least squares solutions form a convex set?".
The derivative is
$$
\begin{align}
  \frac{d}{d\mu} \sum_{k=1}^{m} \left( x_{k} - \mu \right)^{2} &= - 2 \sum_{k=1}^{m} \left( x_{k} - \mu \right) 
\\ &= -2 \left ( \sum_{k=1}^{m} x_{k} - \mu \sum_{k=1}^{m} 1 \right )
\\ &= -2 \left ( \sum_{k=1}^{m} x_{k} - m \mu \right )
\end{align}
\tag{2}
$$
Using the results of $(2)$ in $(1)$ produces the answer
$$
 m \mu = \sum_{k=1}^{m} x_{k} 
\qquad \Rightarrow \qquad
\boxed{
\mu = \frac{1}{m} \sum_{k=1}^{m} x_{k} 
}
$$
The answer is the average best typifies a set of test scores.
Not surprising, but revealing.

Example
Sample data
$$
\begin{array}{cc}
 k & x\\\hline
 1 & 81 \\
 2 & 11 \\
 3 & 78 \\
 4 & 18 \\
 5 & 24 \\
\end{array}
$$
Solution
The merit function, the target of minimization, is
$$
\begin{align}
 \sum_{k=1}^{m} \left( x_{k} - \mu \right)^{2} &= (11-\mu )^2+(18-\mu )^2+(24-\mu )^2+(78-\mu )^2+(81-\mu )^2 
\\
&= 5 \mu ^2-424 \mu +13666
\end{align}
$$
Minimizing this function  of $\mu$ would not give you a moment's hesitation. 
$$
\frac{d}{d\mu}\left(5 \mu ^2-424 \mu +13666\right) = -424 + 10 \mu = 0
$$
The answer is the average
$$
  \mu = \frac{\sum_{k=1}^{m} x_{k}}{m} = \frac{212}{5} = 42.4
$$
Visualization
The figure on the left shows the scores for students $1-5$, with the average a dashed line. The right panel shows equation $(1)$ and how it varies with $\mu$. Hopefully, this panel illustrates why you are looking for $0$s of the first derivative.

Notice that the sum of the squares of the errors is not $0$. The sum of the squares of the errors takes the minimum value of $4677.2$ when $\mu = 42.4$.
In summary, step back from the the linear regression case, and look at this example as a problem in calculus.

Final question
Your final question 

Why does a coefficient like a or b become so important? Why and how is a coefficient so prominently related to error? How does it affect anything?

opens another door to deep insight. Let's defer that answer to a new question like How stable are least squares solutions against variations in the data?
A: When we do ordinary least squares, we are, as you said, trying to minimize the sum of the squared errors our approximation makes:
$$Y_i - (\hat{a}x_i + \hat{b}),$$
for choices $\hat{a}, \hat{b}$.  This the (smooth) objective function we are seeking to minimize.  The taking of the derivative is because a necessary condition for a smooth function to attain a minimum is that its derivative is zero.
A: The sum of squared errors $S$ is a function of the coefficients $a$ and $b$. We have
$$S(a,b)=\sum_{i=1}^n{\big(y_i-(ax_i+b)\big)}^2$$
where $(x_i,y_i)$ are given; they are the data points you have beforehand. The only variables here are $a$ and $b$ (which determine the approximating line).
Now, it's a fact of calculus that local extrema (maximum or minimum) of differentiable functions on open sets must have vanishing derivative. In particular, at the global minimum, we must have
$$\nabla S(a,b)=\left(\frac{\partial S}{\partial a}(a,b),\frac{\partial S}{\partial b}(a,b)\right)=0$$
Now, can you see why a global minimum for $S$ exists?
