Solution:

a. One possible arrangement in the $4 \times 4$ case is the following.

b. Let us first show that $S(n) \leq 4 n$. Suppose for contradiction that $S(n)>4 n$ for some value of $n$. Then there exists at least one row in which there are at least 5 sentinels. Among these 5 sentinels, at least 3 look horizontally in the same direction. But then the last point of the row in that direction will be watched over by at least 3 sentinels, which contradicts the hypothesis.

Let us now show that $S(n) \geq 3 n$. For this we generalize the construction of the $4 \times 4$ case by distinguishing two cases.

- Case $n$ even. We show that, for a generic even $n$ of the form $n=2 k$, it is possible to place $4 n-4$ sentinels (and this number is greater than $3 n$ as soon as $n \geq 4$).
We divide the grid into 4 quadrants of size $k \times k$. In the top-left quadrant we place all the sentinels on the points of the diagonal (the one containing the vertex of the initial grid) and on the points just below the diagonal. All these sentinels have their gaze directed upward and to the left. We place the sentinels in the other 3 quadrants by suitably reflecting the configuration just described.
Note that in each quadrant we have placed exactly $k+(k-1)=n-1$ sentinels, so in total there are $4 n-4$ sentinels.
Note that every point is watched over only by sentinels that are in its quadrant, because the other sentinels look elsewhere. Within the top-left quadrant, a point is watched over only by sentinels that are in the same column and below that point, or in the same row and to the right of that point. If there are no sentinels on a point, these are either only below (and there are at most 2) or only to the right (and there are at most 2). The points on which the sentinels stand are watched over only by the sentinel present there and, at most, by the sentinel just below or just to the right, and thus in every case the sentinels watching over it are at most 2.

- Case $n$ odd. We show that for a generic odd $n$ of the form $n=2 k-1$, with $k \geq 2$, it is possible to place $4 n-3$ sentinels (and this number is greater than $3 n$ as soon as $n \geq 3$).
In this case we think of the grid as a union of four rectangles $k \cdot(k-1)$, plus a small central square. In each rectangle we essentially repeat the arrangement of sentinels described for the $k \times k$ blocks of the even case. The condition that every point is watched over by at most two sentinels is satisfied for the same reason as in the even case.
In this way we have placed $2(k-1)$ sentinels in each rectangle, for a total of $8 k-8=4 n-4$ sentinels. We now note that in the central row and column we have placed only two sentinels, so we can still place one further sentinel in the small central square, with its gaze directed in any two directions we like, without violating the conditions of the problem.

The figure below illustrates the constructions described above in the particular cases $n=10$ and $n=11$. In the $11 \times 11$ case we can orient the sentinel in the central cell as we like.