IIUC，这里有一个可能的选择：

from itertools import permutations

s = (df["Age range"].str.split("-").apply(
    lambda x: pd.Interval(*map(int, x), closed="both")))

d = {l:r for l,r in permutations(s, r=2) if l in r} # or l.overlaps(r) ?

out = (s.map(d).fillna(s).map(lambda x: f"{int(x.left)}-{int(x.right)}")
       .value_counts().reindex(df["Age range"], fill_value=1).reset_index())

输出量：

print(out)

  Age range  count
0      5-12      2
1     12-18      1
2     11-14      1
3       7-9      1
4     17-22      1

conditional_join提供了一种处理不等式连接的有效方法：

# pip install pyjanitor
import pandas as pd
import janitor

splitter = (df['Age range']
            .str
            .split('-',expand=True)
            .set_axis(['left','right'],axis=1)
            .astype(int)
)
out = df.assign(**splitter)
out = (out
       .conditional_join(
           out.assign(counter=1), 
           ('left', 'left', '<'), 
           ('right', 'right', '>'), 
           how = 'left', 
           df_columns = ['Name', 'Age range'],
           right_columns='counter')
      .fillna({'counter':0})
      .assign(counter = lambda f: f.counter + 1)
     )

out = out.assign(counter = out.groupby('Age range').counter.transform('sum'))

print(out)

  Name Age range  counter
0    A      5-12      2.0
1    B     12-18      1.0
2    C     11-14      1.0
3    D       7-9      1.0
4    E     17-22      1.0