Total unimodularity means that every square submatrix of your system matrix has determinant zero, one, or minus one.
Cramer's rule states that the $\lambda_j$ which solve the system $k_{r+1}=\sum_{j=1}^r \lambda_j k_j$ (not that I'm writing $r+1$ instead of $m$, as I find that a little bit clearer) can be written as ratio's of determinants of square submatrices of the system matrix; this is a general property of non-singular linear systems.
But, why is the system $k_{r+1}=\sum_{j=1}^r \lambda_j k_j$ non-singular? In this case, the non-singularity follows from the implicit assumption that the $k_1, \dots, k_r$ are linearly independent (even though $k_1, \dots, k_r, k_{r+1}$ are not). Indeed, if vectors $k_1,\dots,k_{r+1}$ are linearly dependent, then it is always possible to find a subset of vectors such that $k_1,\dots,k_{r'}$ are linearly independent but $k_1,\dots,k_{r'+1}$ are linearly dependent. To see this, start with the first vector $k_1$, and keep adding vectors, stopping as soon as you have a linearly dependent set of vectors.
I agree that your textbook muddles this point a little bit as it does not make it clear that the $k_1, \dots, k_r$ can be assumed linearly independent.
So, from this we know that the $\lambda_i$ are all $0$, $+1$, or $-1$. Without loss of generality, by dropping any columns with coefficients $0$, we know that there are columns in the set of fields such that:
$$\sum_{j=1}^{r+1}\lambda_j k_j=0$$
with all $\lambda_j\in\{+1,-1\}$. I've not reflected the change of indexing when removing columns that have zero coefficients, for ease of notation.
Now consider some column $k_{\ell}$ with $\ell\in\{1,\dots,r+1\}$, say with coefficient $\lambda_{\ell}=+1$, and with $k_{\ell i}=1$ and $k_{\ell j}=1$, where $i$ corresponds to a specific row in the transportation tableau, and $j$ corresponds to a specific column in the transportation tableau (and $k_{\ell}$ is zero everywhere else except at these two indices). By the linear relation that we derived, there must be at least one further column $k_{\ell'}$ which has a coefficient $\lambda_{\ell'}=-1$ and which also has $k_{\ell'i}=1$, and thus, corresponding to the same row in the transportation tableau. Similarly, there must be at least one further column $k_{\ell''}$ which has a coefficient $\lambda_{\ell''}=-1$ and which also has $k_{\ell''j}=1$, and thus, corresponding to the same column in the transportation tableau.
In case $\lambda_\ell=-1$, the reasoning is the same, except that now we'll have coefficients $\lambda_{\ell'}=+1$ and $\lambda_{\ell''}=+1$.
Remember that each column corresponds to a field in the transportation tableau. So, for every field in the set, we have shown that there is another field in the set on the same row of the tableau, and another field in the set on the same column of the tableau. Whence, these fields must clearly form a cycle.
To wrap up, we have shown that, for every set of fields, if the columns of the system matrix corresponding to the set are linearly dependent, then the set of fields must contain a cycle. Note that there may be more fields on the set than just the ones in the cycle. This makes sense, since we can always add fields and obviously still retain linear dependence.
Best Answer
The rows represent the supply and demand nodes. The columns represent the decision variables $x_{ij}$; i.e., how much you send from supply node $i$ to demand node $j$. That's why there are $m+n$ rows and $mn$ columns. The matrix itself is the coefficient matrix for the constraints in the (balanced) transportation problem; in other words, the left-hand side coefficients of the constraints $$\sum_{j=1}^n x_{ij} = s_i, \text{ for each supply node $i$;}$$ $$\sum_{i=1}^m x_{ij} = d_j, \text{ for each demand node $j$}.$$ These come from the assumption that all supply is shipped and all demand is met.
There's a redundant constraint: If you know that all but one of the supply and demand constraints have been satisfied, then the other one must be satisfied as well in order for all supply to be shipped and all demand to be met. In linear algebra terminology, this means that the constraint matrix has rank $m+n-1$ or that it has $m+n-1$ linearly independent columns. We are interested in this because this means that there are $m+n-1$ basic variables in each iteration of the simplex method rather than the $m+n$ basic variables we would normally get with a constraint matrix that has $m+n$ rows and $mn$ columns.
With respect to the stepping stone method, this means that there are only $m+n-1$ circled numbers in the tableau (i.e., flows in the basis, or basic variables) rather than the $m+n$ one would normally expect. It doesn't affect the loops generated or the net effects of the loops, however. If you use the modified distribution method, though, this also means that one of the $m + n$ dual variables $u_i$ and $v_j$ is a free variable. So you can choose a value for one of those variables that makes the subsequent calculations of $c_{ij} - u_i - v_j$ for each nonbasic variable easier. It's common, as in the example in the link, to choose one of $u_i$ or $v_j$ to be $0$.