Let $\textbf{y} = [Y_1, \dotsc, Y_n]^T$ and $\textbf{y}^* = [Y_1 - \overline{Y}, \dotsc, Y_n - \overline{Y}]^T$. Apparently, by subtracting the mean from each of its coordinates, there is linear restriction on $\textbf{y}^*$, ensuring that $\sum(Y_i - \overline{Y}) = 0$. The sum is zero because $$\sum(Y_i - \overline{Y}) = \sum Y_i - n\overline{Y} = n\overline{Y} - n\overline{Y} = 0.$$ $\textbf{Question:}$ Among the $n$ values of $Y_i - \overline{Y}$, why only $n-1$ are linearly independent?
If I understand this, then I know why the total sum of squares TSS $= ||\textbf{y}^*||^2 = \sum(Y_i - \overline{Y})^2$ has $n-1$ degrees of freedom. This is because degrees of freedom associated with a sum of squares represent the dimension of the subspace to which the corresponding vector is confined.