Let's consider a rigid body B and an arbitrary reference point P ${\it fixed}$ with respect to the body (P can be a particle of the body or it can be a "mathematical point" outside the body but solidary to B, it doesn't matter). As B moves, the point P has a velocity ${\bf v}_P$ that, of course, can change over time. The most general motion of B is the composition of a rigid translation with velocity ${\bf v}_P$ and a rotation with angular velocity $\vec \omega$ around an axis that passes through the point P (This result is known as the Chasles' theorem). Accordingly, the velocity of any point A of the body is
\begin{equation}{\bf v}_A = {\bf v}_P + {\vec \omega}\wedge \left( {\bf r}_A - {\bf r}_P\right),\;\;\;\;\;\;\;\;\;\;\;\; (1)
\end{equation}
where ${\bf r}_A$, ${\bf r}_P$ are the vector position of the points A and P with respect to a given reference frame $S.$ This is a kinematic result, it doesn't matter if the reference frame is non-inertial. When we compute the kinetic energy with respect to $S$, $$T=\frac{1}{2}\sum_{i\in B}m_i {\bf v}_i^2,$$
using Eq. (1), $T$ results the sum of three terms: one involving only the rigid translation (${\bf v}_P$), other involving only the rotation ($\vec\omega$), and a third one that mixes translation and rotation. In the special case where the center of mass is taken as the point P, the last term vanishes and, then, the kinetic energy is the sum of the translational kinetic energy (the body moving as a whole with the center of mass velocity) and the rotational kinetic energy (the body rotating around its center of mass):
$$T= \frac{1}{2}M{\bf v}_{CM}^2 + \frac{1}{2}\sum_{i \in B}
m_i \left(\vec \omega \wedge ({\bf r}_i-{\bf r}_{CM})\right)^2.\;\;\;\;(2)$$
In the case of a rotation around a fixed direction (like in your problem), the last term is rewritten as
$$T_{rot} = \frac{1}{2}I^{cm}\omega^2,$$
where $I^{cm}$ is the body moment of inertia with respect to the rotation axis that pass through the center of mass of the body.
If one point of the body is fixed with respect to $S,$ all the kinetic energy has a rotational character (see Eq. (1), with ${\bf v}_P={\bf 0}$),
$$T = T_{rot} = \frac{1}{2}I^{P}\omega^2,\;\;\;\;(3)$$
where now $I^{P}$ is the body moment of inertia with respect to an axis that pass through the point P (again we are considering rotation around a fixed direction).
Let's go to your problem. With respect to the rod, we can take the pivot as the reference point $P$ (this is convenient, as the pivot is fixed with respect to a laboratory frame). So, the kinetic energy of the rod is (see Eq. (3)),
$$T_{rod}= \frac{1}{2}I_{rod}^P \dot{\theta}^2,$$ where, for a uniform rod $I^P_{rod} = \frac{1}{3}M_{rod} L_{rod}^2,$ and $\theta$ is the angle between the rod and the vertical direction (You can recalculate the rod kinetic energy taking its center of mass as the reference point $P$!).
With respect to the square block, it is convenient to take its center of mass as the reference $P$ point. In this case, the velocity of the center of mass is $|{\bf v}^{cm}_{block}| = L_{rod} |\dot\theta|$, while its angular velocity will depend on whether this block is allowed to rotate independently of the rod, or not. In the first case, if we call $\phi$ the angle that determines the block orientation, the total block kinetic energy is (see Eq. (2)),
$$T_{block} = \frac{1}{2}M_{block} L_{rod}^2 \dot\theta^2 + \frac{1}{2}I^{cm}_{block}\dot\phi^2.$$
If the square block is rigidly tied to the rod (that is, it cannot rotate independently), its angular velocity will be the same as the one for the rod ($\dot \theta$). So, in the last equation you should replace $\dot\phi$ by $\dot\theta$.
Don't forget to include the potential energy in the Lagrangean function.
