How does the Nearest Centroid method work?
You are making a mistake regarding what is given: during training, you don't have a radius $R$. You have the coordinates $\vec{x}$ and the label $y$ for each point:
$$ \vec{x}_1 = [-2;1] \quad \vec{x}_2 = [1;2] \quad \vec{x}_3 = [0;-1] \quad \vec{x}_4 = [1;0] \quad \vec{x}_5 = [1;1] $$
and
$$ y_1 = T \quad y_2=T \quad y_3=T \quad y_4=S \quad y_5=S $$
with that, your "trained" centroids are
$$ \mu_T = \frac 13 [-2 + 1 + 0; 1 + 2 - 1] = \left[-\frac 13, \frac 23\right] $$
$$ \mu_S = \frac 12 [1 + 1; 0 + 1] =\left[1, \frac 12\right]$$
You calculate these centroids before you get any test values.

Then, for testing, for your observed point $\vec{x} = [1,1]$, you calculate the Euclidean distance between the point $\vec{x}$ and the centroids $\mu_T$ and $\mu_S$:
$$ \| \vec{x} - \mu_T \| = \left\| [1;1] - \left[-\frac 13; \frac 23\right] \right\| = \left\| \left[\frac 43; \frac 13\right] \right\| = 1.37 $$ 
$$ \| \vec{x} - \mu_S \| = \left\| [1;1] - \left[1; \frac 12\right] \right\| = \left\| \left[0; \frac 12 \right] \right\| = 0.5$$
Finally, the term $\hat{y} = \arg\min_{l \in \mathbf{Y}} \|\vec{x}-\mu_l\|$ is used to find the estimated class $\hat{y}$ (the hat symbol is to denote that this is an estimated $y$, not one we knew before.). $\arg\min$, means that you find the minimum value - which is 0.5 in our case, and chose which "argument", i.e. which class, leads to that minimum value. In our case, the class which leads to the minimal distance is $S$, so the result is $\hat{y} = S$, and our test point is a square.