The quadrilateral $K B C L$ touches all its sides to the inscribed circle of $\triangle A B C$. At each vertex, there are two equal tangent segments. Two opposite sides of the quadrilateral consist precisely of the four different tangent segments, so $|K B|+|L C|=|K L|+|B C|$.

Given $K L \| B C$, we have $\triangle A K L \sim \triangle A B C$. Let $t$ be the scaling factor of this similarity, so $|A K|=t|A B|,|A L|=t|A C|$, and $|K L|=t|B C|$. Then $|B K|=$ $|A B|-|A K|=|A B|-t|A B|=(1-t)|A B|$. Similarly, $|C L|=(1-t)|A C|$. Now we have
$t|B C|+|B C|=|K L|+|B C|=|K B|+|L C|=(1-t)|A B|+(1-t)|A C|=3(1-t)|B C|$,
where the last equality holds due to the given condition in the problem. Dividing by $|B C|$ gives $1+t=3-3 t$, or $4 t=2$ thus $t=\frac{1}{2}$. We conclude that $K$ is the midpoint of $A B$ and $L$ is the midpoint of $A C$.

We already knew $\triangle A K L \sim \triangle A B C$, which implies $\frac{|K L|}{|B C|}=\frac{|A L|}{|A C|}$. Given $K L \| B C$, we also have $\triangle T S L \sim \triangle T B C$, so $\frac{|S L|}{|B C|}=\frac{|T L|}{|T C|}$. Combining these ratios, we get

$$\frac{|S L|}{|K L|}=\frac{|S L|}{|B C|} \cdot \frac{|B C|}{|K L|}=\frac{|T L|}{|T C|} \cdot \frac{|A C|}{|A L|} .$$

We know that $\frac{|A C|}{|A L|}=2$, since $L$ is the midpoint of $A C$. This also means that $T$ is the midpoint of $A L$, as $4|A T|=|A C|=2|A L|$. Thus $\frac{|T L|}{|T C|}=\frac{\frac{1}{4}|A C|}{\frac{3}{4}|A C|}=\frac{1}{3}$. We conclude that $\frac{|S L|}{|K L|}=\frac{1}{3} \cdot 2=\frac{2}{3}$.