Q:

Intuition behind regression sum of squares

$SSR = \sum_{i = 1}^{n} (\hat{Y}_i - \bar{Y})^2$ is the sum of squares of the difference between the fitted value and the average response variable. In other words, it measures how far the regression line is from $\bar{Y}$. Higher $SSR$ leads to higher $R^2$, the coefficient of determination, which corresponds to how well the model fits our data. I'm having trouble wrapping my mind around why the farther away the regression line is from the average $Y$ means that the model is a better fit. 

A:

Just a bit of a misunderstanding with the definitions, I believe:
\begin{align}
\text{SST}_{\text{otal}} &= \color{red}{\text{SSE}_{\text{xplained}}}+\color{blue}{\text{SSR}_{\text{esidual}}}\\
\end{align}
or, equivalently,
\begin{align}
\sum(y_i-\bar y)^2 &=\color{red}{\sum(\hat y_i-\bar y)^2}+\color{blue}{\sum(y_i-\hat y_i)^2}
\end{align}
and
$\large \text{R}^2 = 1 - \frac{\text{SSR}_{\text{esidual}}}{\text{SST}_{\text{otal}}}$
So if the model explained all the variation, $\text{SSR}_{\text{esidual}}=\sum(y_i-\hat y_i)^2=0$, and $\bf R^2=1.$
From Wikipedia:

Suppose $r = 0.7$ then $R^2 = 0.49$ and it implies that $49\%$ of the
  variability between the two variables have been accounted for and the
  remaining $51\%$ of the variability is still unaccounted for.

The sum of the squared distances between the mean ($\bar Y$) and the fitted values ($\hat Y$) (the SSExplained) is the part of the distance from the mean to the actual value ($ Y$) (TSS) that the model has been able to account for. The difference between these two calculations, is the unexplained part of the variation (the residuals). If you take TSS as a fixed value, the higher the SSExplained, the lower the SSResidual, and hence the closer to 1 R.Square will be.

Here is some intuition, at the risk of actually making clear waters murky. In OLS we minimize distances to the points in the data cloud in an overdetermined system, rendering a line that fulfills $\text{SST}>\text{SSE}$. The difference is the $\text{SSR}$ (residuals).
But let's imagine a data "cloud" of three points, all perfectly aligned. Now, let's play a game of actually doing the opposite of an OLS: we are going to increase the error by proposing a line different from the line that goes through all the points, using the mean as a fulcrum. Remember that the OLS goes through the mean values $({\bf \bar X, \bar Y})$, which is the blue point in the middle, through which we draw a horizontal line. In this case, opposite to the expected situation in OLS and just to illustrate the point, we can see how by moving the line from having zero $\text{SSR}$ (all the variance, $\text{SST}$ accounted by the model (the line), $\text{SSE}$) on the left "column" of the diagram, we introduce residual errors (in red, on the right part of the diagram):

Logically, by minimizing errors, and in the typical situation of an overdetermined system, the $\text{SST}> \text{SSE}$, and the difference will correspond to the $\text{SSR}$.

Here is a quick example with a widely available data set in R:
fit = lm(mpg ~ wt, mtcars)
summary(fit)$r.square
[1] 0.7528328
> sse = sum((fitted(fit) - mean(mtcars$mpg))^2)
> ssr = sum((fitted(fit) - mtcars$mpg)^2)
> 1 - (ssr/(sse + ssr))
[1] 0.7528328

