Variance of ridge regression estimator

Let $U\Lambda U'=X'X$ be the eigendecomposition of $X'X$, where $\Lambda$ is diagonal and $U$ is orthogonal. Then $X'X + kI = U(\Lambda + kI)U'$, and thus $$\sigma^2 (X'X + kI)^{-1} X'X (X'X + kI)^{-1} = \sigma^2 U(\Lambda + kI)^{-1} \Lambda (\Lambda + kI)^{-1} U',$$ where we have used the fact that $U$ is orthogonal.

$\sum_i \text{Var}((\hat{\beta}_R)_i)$ is the trace of the covariance matrix $\text{Var}(\hat{\beta}_R)$. Using the linearity and cyclic properties of trace along with orthogonality of $U$, $$\operatorname{Tr} (\sigma^2 U(\Lambda + kI)^{-1} \Lambda (\Lambda + kI)^{-1} U') = \sigma^2 \operatorname{Tr}((\Lambda + kI)^{-1} \Lambda (\Lambda + kI)^{-1}).$$ The expression inside the trace is a diagonal matrix with entries $\frac{\lambda_i}{(\lambda_i+k)^2}$.