using cartesian coordinates as an intermediate step, the kinetic energy is calculated as such
why is it incorrect to just say that the kinetic energy for the first bob is $$T_1 = (1/2) m_1 (l_1\dot{\theta_1})^2$$
and for the second bob is $$T_2 = (1/2) m_2 (l_2\dot{\theta_2}+l_1 \dot{\theta_1}))^2$$
where is the $cos(\theta_1 - \theta_2)$ term coming from?
The expression $T_2 = (1/2) m_2 (l_2\dot{\theta_2}+l_1 \dot{\theta_1}))^2$ is incorrect because you are just adding up two vector quantities algebraically.
The vector form the kinetic energy of a particle with mass $m$ is
$$ T = \tfrac{1}{2} m (\vec{v} \cdot \vec{v}) = \tfrac{1}{2} m \| \vec{v} \|^2 $$
The expression you are asking about is equal to
$$ T \overset{?}{=} \tfrac{1}{2} m_1 \| \vec{v}_1 \|^2 + \tfrac{1}{2} m_2 \| \vec{v}_2 - \vec{v}_1 \|^2 $$
which is obviously not equal to the correct expression
$$ T = \tfrac{1}{2} m_1 \| \vec{v}_1 \|^2 + \tfrac{1}{2} m_2 \| \vec{v}_2 \|^2 $$
Your first analysis is correct because it splits the velocity vector into components and considers the correct combination of $\dot{\theta}_1$ and $\dot{\theta}_2$ that yields kinetic energy.
