A regular tetrahedron is a three-dimensional shape with four faces, each of which is an equilateral triangle. All edges are of equal length, in this case, 4 cm.

1. Draw an equilateral triangle PQR with side length 4 cm.

2. Draw another equilateral triangle SQR, sharing the edge QR, such that vertex S is on the opposite side of QR from vertex P. This forms a larger triangle (or a rhombus if P, Q, S, R were coplanar, but here P and S are vertices of two triangles sharing a base).

   The net will look like this:

         P
        / \
       /   \
      /     \
     Q-------R
      \     /
       \   /
        \ /
         S
   

   Copy

1. Mark the midpoint T on the edge QR.
2. The straight path from P to S, passing through T, is a straight line segment connecting P and S on this unfolded net.

1. The line segment PT is the altitude (height) of the equilateral triangle PQR from vertex P to the base QR. The formula for the altitude $h$ of an equilateral triangle with side length $a$ is $h = \frac{\sqrt{3}}{2}a$. For $\triangle PQR$, $a = 4$ cm. So, the length of $PT = \frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3}$ cm.

2. Similarly, the line segment ST is the altitude of the equilateral triangle SQR from vertex S to the base QR. For $\triangle SQR$, $a = 4$ cm. So, the length of $ST = \frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3}$ cm.

3. On the unfolded net, P, T, and S are collinear. Therefore, the total length of the straight path PS is the sum of PT and ST. Length of path PS = $PT + ST = 2\sqrt{3} \text{ cm} + 2\sqrt{3} \text{ cm} = 4\sqrt{3} \text{ cm}$.

4. To get a numerical value, we use $\sqrt{3} \approx 1.732$. Length of path PS $\approx 4 \times 1.732 = 6.928$ cm.

The length of the straight path of PS through the midpoint T over the edge QR is $\boxed{4\sqrt{3} \text{ cm}}$ or approximately $\boxed{6.93 \text{ cm}}$.