The cross-terms of the generalized inertia tensor $\mathbf{M}$ are zero, $M_{r\varphi} = M_{\varphi r} = 0$, precisely because the total kinetic energy ($T$) contains no mixed product terms of the generalized velocities, i.e., no $\dot{r}\dot{\varphi}$ term.
Here is the explanation for why this is the case, based on the definition of the components:
The components of the generalized inertia tensor are defined by the second partial derivatives of the kinetic energy ($T$) with respect to the generalized velocities:
$$ M_{ij} = \frac{\partial^2 T}{\partial \dot{q}_i \partial \dot{q}_j} $$
For the specific cross-term $M_{r\varphi}$ (where $\dot{q}_i = \dot{r}$ and $\dot{q}_j = \dot{\varphi}$), the calculation is:
$$ M_{r\varphi} = \frac{\partial^2 T}{\partial \dot{r} \partial \dot{\varphi}} $$
The kinetic energy ($T$) for the system described on the web page is:
$$ T = \frac{1}{2} (m_1 + m_2) \dot{r}^2 + \frac{1}{2} (m_1 r^2) \dot{\varphi}^2 $$
First Partial Derivative ($\frac{\partial T}{\partial \dot{\varphi}}$):
We take the derivative of $T$ with respect to the angular velocity $\dot{\varphi}$:
$$ \frac{\partial T}{\partial \dot{\varphi}} = \frac{\partial}{\partial \dot{\varphi}} \left[ \frac{1}{2} (m_1 + m_2) \dot{r}^2 + \frac{1}{2} (m_1 r^2) \dot{\varphi}^2 \right] $$
The first term, $\frac{1}{2} (m_1 + m_2) \dot{r}^2$, is independent of $\dot{\varphi}$, so its derivative is 0.
The derivative of the second term is $\frac{1}{2} (m_1 r^2) (2\dot{\varphi}) = (m_1 r^2) \dot{\varphi}$.
$$ \frac{\partial T}{\partial \dot{\varphi}} = m_1 r^2 \dot{\varphi} $$
Second Partial Derivative ($M_{r\varphi}$):
Now, we take the derivative of the result from step 1 with respect to the radial velocity $\dot{r}$:
$$ M_{r\varphi} = \frac{\partial}{\partial \dot{r}} \left[ m_1 r^2 \dot{\varphi} \right] $$
Since the entire term $m_1 r^2 \dot{\varphi}$ is treated as a constant with respect to $\dot{r}$ (it does not contain $\dot{r}$), its derivative is zero.
$$ M_{r\varphi} = 0 $$
Symmetry ($M_{\varphi r}$):
Since the order of differentiation does not matter for well-behaved functions (which $T$ is), the matrix $\mathbf{M}$ is symmetric, meaning $M_{\varphi r} = M_{r\varphi} = 0$.