Denote the radius of the incircle of $\triangle ABC$ by $r$. Then the area of triangle $ABC$ is
$$\frac{1}{2}|AB| \cdot r + \frac{1}{2}|BC| \cdot r + \frac{1}{2}|AC| = \frac{1}{2}r \cdot (3|BC| + |BC|) = 2r|BC|.$$
On the other hand, the area of $ABC$ equals $\frac{1}{2}h|BC|$, where $h$ is the altitude from $A$. Hence, $h = 4r$. Because the distance from $KL$ to $BC$ is exactly $2r$, the distance from $A$ to $KL$ is also $2r$. Triangles $AKL$ and $ABC$ are similar, because $KL \parallel BC$, and the altitudes from $A$ have lengths $2r$ and $4r$, respectively, giving a multiplication factor of exactly 2. Hence, $K$ is the midpoint $AB$, and $L$ is the midpoint of $AC$.
For the point $T$, we have $|AC| = 4|AT|$, hence $|AT| = \frac{1}{4}|AC| = \frac{1}{2}|AL|$, hence $T$ is the midpoint of $AL$. Now consider triangle $ABL$. In this triangle, the segment $BT$ is a median, because $T$ is the midpoint of $AL$. Also $LK$ is a median as $K$ is the midpoint $AB$. Their intersection point $S$ is the centroid, from which we get that $\frac{|SL|}{|KL|} = \frac{2}{3}$.
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