= Solution
The two positions are $\mathbf r_1=(2l\sin\theta_1,-2l\cos\theta_1)$ and $\mathbf r_2=\mathbf r_1+(l\sin\theta_2,-l\cos\theta_2)$. Differentiating and summing their <kinetic energies> gives the exact expression
$$
K=\frac{ml^2}{2}[8\dot\theta_1^2+\dot\theta_2^2+4\dot\theta_1\dot\theta_2\cos(\theta_1-\theta_2)].
$$
In the linear small-oscillation approximation, set the cosine to one in this quadratic velocity expression:
$$
\boxed{K\simeq\frac{ml^2}{2}(8\dot\theta_1^2+4\dot\theta_1\dot\theta_2+\dot\theta_2^2).}
$$
The cross term arises because the lower mass participates in motion generated by both rods.
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