Extração do Oracle XMLTYPE com base no valor e na condição

Você pode voltar até o irmão do aluno s_days nó:

select h.PlanCodeCode, b.amount, b.pcode, b.child1_amount, b.child2_amount
 from   t
    cross join
    xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             '/SSO_XML'
             passing t.xml
             columns PlanCodeCode varchar2(100)  path './PlanCode/@PlanCodeCode',
                     attributes xmltype path './PlanCode'
            ) h
    left join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns node_level for ordinality
                    , amount number path '@Amount'
                    , pcode  varchar2(10) path './../../@PCODE'
                    , child1_amount number path './../../../S_DAYS[@PCODE="Child1"]/AdditonalFare/AdditonalFareAmount/@Amount'
                    , child2_amount number path './../../../S_DAYS[@PCODE="Child2"]/AdditonalFare/AdditonalFareAmount/@Amount'
            ) b on 1=1;

Ou você pode obter os filhos da primeira XMLTable, se sempre quiser vê-los, mesmo que não haja nós de alunos:

select h.PlanCodeCode, b.amount, b.pcode, h.child1_amount, h.child2_amount
 from   t
    cross join
    xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             '/SSO_XML'
             passing t.xml
             columns PlanCodeCode varchar2(100)  path './PlanCode/@PlanCodeCode',
                     attributes xmltype path './PlanCode',
                     child1_amount number path './PlanCode/S_DAYS[@PCODE="Child1"]/AdditonalFare/AdditonalFareAmount/@Amount',
                     child2_amount number path './PlanCode/S_DAYS[@PCODE="Child2"]/AdditonalFare/AdditonalFareAmount/@Amount'
            ) h
    left join xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns node_level for ordinality
                    , amount number path '@Amount'
                    , pcode  varchar2(10) path './../../@PCODE'
            ) b on 1=1;

Aliás, como você está no 12c, você pode usar cross apply e outer apply - o último em vez da junção externa com on 1=1 fictício doença.

select h.PlanCodeCode, b.amount, b.pcode, h.child1_amount, h.child2_amount
 from   t
    cross apply
    xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             '/SSO_XML'
             passing t.xml
             columns PlanCodeCode varchar2(100)  path './PlanCode/@PlanCodeCode',
                     attributes xmltype path './PlanCode',
                     child1_amount number path './PlanCode/S_DAYS[@PCODE="Child1"]/AdditonalFare/AdditonalFareAmount/@Amount',
                     child2_amount number path './PlanCode/S_DAYS[@PCODE="Child2"]/AdditonalFare/AdditonalFareAmount/@Amount'
            ) h
    outer apply xmltable(xmlnamespaces(default 'http://www.w3.org/2001/XMLSchema'),
             'PlanCode/S_DAYS/STUDENT/DIVISION'
             passing h.attributes
             columns node_level for ordinality
                    , amount number path '@Amount'
                    , pcode  varchar2(10) path './../../@PCODE'
            ) b;

Qualquer um deles obtém o mesmo resultado com seus dados de amostra:

PLANCODECODE | AMOUNT | PCODE | CHILD1_AMOUNT | CHILD2_AMOUNT
:----------- | -----: | :---- | ------------: | ------------:
CHOICE       | 150.05 | P123  |           100 |           130
CHOICE       | 250.05 | P123  |           100 |           130
CHOICE       | 150.05 | P1234 |           100 |           130
CHOICE       | 250.05 | P1234 |           100 |           130

db<>fiddle