$J$ is the result of applying a sequence of row operations to $I_m$, and therefore $J$ and $I_m$ are row-equivalent. $\homosystem{I_m}$ has only the zero solution, since $I_m$ is nonsingular (Theorem NMRRI). Thus, $\homosystem{J}$ also has only the zero solution (Theorem REMES, Definition ESYS) and $J$ is therefore nonsingular (Definition NSM).
To prove the second part of this conclusion, first convince yourself that row operations and the matrix-vector product are associative operations. By this we mean the following. Suppose that $F$ is an $m\times n$ matrix that is row-equivalent to the matrix $G$. Apply to the column vector $F\vect{w}$ the same sequence of row operations that converts $F$ to $G$. Then the result is $G\vect{w}$. So we can do row operations on the matrix, then do a matrix-vector product, or do a matrix-vector product and then do row operations on a column vector, and the result will be the same either way. Since matrix multiplication is defined by a collection of matrix-vector products (Definition MM), the matrix product $FH$ will become $GH$ if we apply the same sequence of row operations to $FH$ that convert $F$ to $G$. (This argument can be made more rigorous using elementary matrices from the upcoming Subsection DM.EM and the associative property of matrix multiplication established in Theorem MMA.) Now apply these observations to $A$.
Write $AI_n=I_mA$ and apply the row operations that convert $M$ to $N$. $A$ is converted to $B$, while $I_m$ is converted to $J$, so we have $BI_n=JA$. Simplifying the left side gives the desired conclusion.
For the third conclusion, we now establish the two equivalences \begin{align*} A\vect{x}&=\vect{y} & &\iff & JA\vect{x}&=J\vect{y} & &\iff & B\vect{x}&=J\vect{y} \end{align*}
The forward direction of the first equivalence is accomplished by multiplying both sides of the matrix equality by $J$, while the backward direction is accomplished by multiplying by the inverse of $J$ (which we know exists by Theorem NI since $J$ is nonsingular). The second equivalence is obtained simply by the substitutions given by $JA=B$.
The first $r$ rows of $N$ are in reduced row-echelon form, since any contiguous collection of rows taken from a matrix in reduced row-echelon form will form a matrix that is again in reduced row-echelon form (Exercise RREF.T12). Since the matrix $C$ is formed by removing the last $n$ entries of each these rows, the remainder is still in reduced row-echelon form. By its construction, $C$ has no zero rows. $C$ has $r$ rows and each contains a leading 1, so there are $r$ pivot columns in $C$.
The final $m-r$ rows of $N$ are in reduced row-echelon form, since any contiguous collection of rows taken from a matrix in reduced row-echelon form will form a matrix that is again in reduced row-echelon form. Since the matrix $L$ is formed by removing the first $n$ entries of each these rows, and these entries are all zero (they form the zero rows of $B$), the remainder is still in reduced row-echelon form. $L$ is the final $m-r$ rows of the nonsingular matrix $J$, so none of these rows can be totally zero, or $J$ would not row-reduce to the identity matrix. $L$ has $m-r$ rows and each contains a leading 1, so there are $m-r$ pivot columns in $L$.