Using the properties of the covariance operator we have
$$$$
$$\begin{align*}
\textrm{Cov}(Y_{ij}, Y_{ik})
& = \textrm{Cov}(\beta_0 + \beta_1 t_{ij} + b_{0i} + b_{1i} t_{ij} + e_{ij}, \;
\beta_0 + \beta_1 t_{ik} + b_{0i} + b_{1i} t_{ik} + e_{ik}) \\
& = \textrm{Cov}(b_{0i} + b_{1i} t_{ij} + e_{ij}, \;
b_{0i} + b_{1i} t_{ik} + e_{ik}) \\
& = \textrm{Cov}(b_{0i}, \; b_{0i})
+ t_{ik} \, \textrm{Cov}(b_{0i}, \; b_{1i})
+ \textrm{Cov}(b_{0i}, \; e_{ik}) \\
& \quad +\, t_{ij} \, \textrm{Cov}(b_{1i}, \; b_{0i})
+ t_{ij} t_{ik} \, \textrm{Cov}(b_{1i}, \; b_{1i})
+ t_{ij} \, \textrm{Cov}(b_{1i}, \; e_{ik}) \\
& \quad +\, \textrm{Cov}(e_{ij}, \; b_{0i})
+ t_{ik} \, \textrm{Cov}(e_{ij}, \; b_{1i})
+ \textrm{Cov}(e_{ij}, \; e_{ik}) \\
& = \textrm{Var}(b_{0i})
+\, (t_{ij} + t_{ik}) \textrm{Cov}(b_{0i}, \; b_{1i})
+\, t_{ij}t_{ik} \textrm{Var}(b_{1i})
+\, \textrm{Cov}(e_{ij}, \; e_{ik})
\end{align*}$$
$$$$
If $j \neq k$, then
$$\begin{align*}
\textrm{Cov}(Y_{ij}, Y_{ik})
& = \textrm{Var}(b_{0i})
+ (t_{ij} + t_{ik}) \, \textrm{Cov}(b_{0i}, \; b_{1i})
+ t_{ij} t_{ik} \textrm{Var}(b_{1i})
\end{align*}$$
If $j = k$, then
$$\begin{align*}
\textrm{Cov}(Y_{ij}, Y_{ik})
& = \textrm{Var}(Y_{ij}) \\
& = \textrm{Var}(b_{0i})
+ 2 \, t_{ij} \, \textrm{Cov}(b_{0i}, \; b_{1i})
+ t_{ij}^2 \textrm{Var}(b_{1i})
+ \sigma^2
\end{align*}$$
$$$$
$\textrm{Var}(b_{0i})$, $\textrm{Var}(b_{1i})$, and $\textrm{Cov}(b_{0i}, \; b_{1i})$ are found in your matrix $D$.