To show that the matrix is singular, it suffices to show that $\det\left(\frac{1}{4} BA^T - I\right) = 0$, i.e. $\lambda = 4$ is an eigenvalue of $BA^T$.
Before we start, we can actually look at the example you provided. Let's try $A = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$. Then $BA^T = \begin{pmatrix} 4 & 3 \\ 0 & 1 \end{pmatrix}$. At this point we can already make a hypothesis / observation that each column sums up to $4$, which is what we will show.
The easiest without any magical intuition is to expand everything. Write $B = (b_{ij})$ and $A^T = (a_{ij})$. By the constraints given, $\sum_i b_{ij} = \sum_i a_{ij} = 2$ for any fixed $j$.
Now, $(BA^T)_{ij} = \sum_{k = 1}^n B_{ik} A^T_{kj}$ by definition. We then sum up columns while fixing $j$:
$$
\sum_{i = 1}^n (BA^T)_{ij} = \sum_{i = 1}^n \sum_{k = 1}^n B_{ik}A^T_{kj} = \sum_{k = 1}^n \left(\sum_{i = 1}^n B_{ik}A^T_{kj}\right) = \sum_{k = 1}^n A^T_{kj}\left(\sum_{i = 1}^n B_{ik}\right) = 2 \cdot 2 = 4
$$
Now, we have a matrix $BA^T$ where every column is a multiple of $4$. How do we prove that $\lambda = 4$ is an eigenvalue? Here we can use a result which states that $\lambda$ is an eigenvalue of $M$ if and only if $\lambda$ is an eigenvalue of $M^T$! Therefore, we can deal with $(BA^T)^T = AB^T$, where each row sums to $4$ instead. The rest is simple, as I claim that $(1, 1, \cdots, 1)$ is an eigenvector of $AB^T$ with eigenvalue $4$. You can prove this yourself.
Hint: Look at what's happening below
$$
\begin{pmatrix} 0.5 & 0.3 & 3.2 \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} ? & ? & ? \end{pmatrix}
$$
Hope this helps!
