Prefix Sums in Mutable 2D Array
There is a relatively straightforward solution where each query and each update can be done in $O(\log^2 n)$ time.  The data structure uses $O(n^2)$ space.

We will have $\lg n$ "granularities" of data structure, one for each power of two $2^m$ such that $1 \le 2^m \le n$.  The data structure for granularity $2^m$ stores the sums

$$\sum_{i=k_0 \cdot 2^m}^{(k_0+1) \cdot 2^m -1} \sum_{j=0}^{l-1} M[i,j]$$

for each $k_0,l$.  This data structure for granularity $2^m$ can in turn be represented using $n/2^m$ balanced trees (one for each possible value of $k_0$) to store prefix sums.

Now to look up the prefix sum for $k,l$, we break up the interval $[0,k-1]$ into a union of intervals of power-of-two length; at most $\lg n$ intervals are needed.  For each such interval of length $2^m$, we do a lookup into the data structure of granularity $2^m$.  Thus queries can be answered by doing $O(\log n)$ lookups into a balanced tree, each of which takes $O(\log n)$ time, for a total time of $O(\log^2 n)$ per query.

Updates can also be done in $O(\log^2 n)$ time.  To update $M[i,j]$, for each granularity $2^m$, you update the appropriate balanced tree in the data structure of granularity $2^m$.  This is $O(\log n)$ updates oin $O(\log n)$ balanced trees; each such takes $O(\log n)$ time, so the total time is $O(\log^2 n)$ time.

Finally, the data structure of granularity $2^m$ contains $n/2^m$ trees, each taking up $O(n)$ space, so the total space usage is $O(n^2 \cdot (1 + 1/2 + 1/4 + \cdots)) = O(n^2)$.