Very quickly -- using the formula $$R^2 (centered) = \dfrac{Model\ Sum\ of\ Squares}{Model\ Sum\ of\ Squares - Residual\ Sum\ of\ Squares} = \dfrac{\sum(\hat y_i - \bar y)^2}{\sum(\hat y_i - \bar y)^2 - \sum(\epsilon_i)^2} $$

With a linear transformation of the dependent variable $y$, the distance between the fitted values and $\bar y$ remains the same. Likewise, since a linear transformation of the dependent variable changes the intercept, but not the slope, the residuals remain the same as well. So, $$MSS = MSS^\prime and\ RSS = RSS^\prime$$ where $MSS^\prime$ and $RSS$ are from the transformed linear model. Therefore, $$R^2(centered) = R^2(centered)^\prime$$

With the uncentered $R^2$ removes the mean of $y$ from the summation. $$R^2 (uncentered) = \dfrac{\sum(\hat y_i)^2}{\sum(\hat y_i)^2 - \sum(\epsilon_i)^2} $$ So, $$MSS \ne MSS^\prime and\ RSS = RSS^\prime$$ Therefore, $$R^2(uncentered) \ne R^2(uncentered)^\prime$$

Very quickly -- using the formula $$R^2 (centered) = \dfrac{Model\ Sum\ of\ Squares}{Model\ Sum\ of\ Squares - Residual\ Sum\ of\ Squares} = \dfrac{\sum(\hat y_i - \bar y)^2}{\sum(\hat y_i - \bar y)^2 - \sum(\epsilon_i)^2} $$

With a linear transformation of the dependent variable $y$, the distance between the fitted values and $\bar y$ remains the same. Likewise, since a linear transformation of the dependent variable changes the intercept, but not the slope, the residuals remain the same as well. So, $$MSS = MSS^\prime and\ RSS = RSS^\prime$$ where $MSS^\prime$ and $RSS$ are from the transformed linear model. Therefore, $$R^2(centered) = R^2(centered)^\prime$$

With the uncentered $R^2$ removes the mean of $y$ from the summation. $$R^2 (uncentered) = \dfrac{\sum(\hat y_i)^2}{\sum(\hat y_i)^2 - \sum(\epsilon_i)^2} $$ So, $$MSS \ne MSS^\prime and\ RSS = RSS^\prime$$ Therefore, $$R^2(uncentered) \ne R^2(uncentered)^\prime$$ The lack of centering for the uncentered $R^2$ means that the value changes after a linear transformation of $y$.