Does the R-Squared value determine the variance in the model for all data points? I understand that the R squared value determines how much of the variability in the response variables is determined by the regression model (I think).
I am confused about whether the R squared variable determines the proportion of variance between any two data points or not.

For example, if we take the two points marked above, will the bit marked x account for 70% of the variability (assuming r squared in 0.7), and will the blue parts be the residual 30%? Or is this not correct?
I am confused as my teacher said that the r squared value determines the variance between two points counted for by the model, but wouldn't this be impossible for every two possible points?
 A: Consider a simple linear regression model $$Y_i = \beta_0 + \beta_1x + e_i,$$ for $i = 1, 2, \dots, n,$ with 
$e_i \sim \mathsf{Norm}(0, \sigma_e).$
Let $S_Y^2$ be the variance of the $Y_i$ and $S_{Y|x}^2$ be the variance of the residuals $r_i = Y_i - \hat Y_i.$
Then
$$S_{Y|x}^2 = \frac{n-1}{n-2}S_Y^2(1 - r^2),$$
where the coefficient of determination $r^2$ is often
printed in computer results as R-sq.
If $r^2 \approx 0,$ then $S_{Y|x}^2 \approx S_Y^2,$
so that the regression is of little value in 'explaining' the variance about the regression line.
By contrast, if $r^2 \approx 1,$ then $S_{Y|x}^2 \approx 0,$ so that most of the $(x_i, Y_i)$ lie on or very near the regression line. This is the basis of the rough statement that "$r^2$ is the proportion of the variance among the $Y_i$'s is explained by regression on $x.$"
The absolute difference between the two values in your graph depends mainly on the slope $\beta_1$ of the true line and on $S_Y^2,$.
As long as $\beta_1 \ne 0,$ slope depends on the units used,  and not directly on the correlation.
