Command grouping (&&, ||, ...)

Operator precedence for && and || is strictly left-to-right.

Thus:

pwd; (ls) || { cd .. && ls student/; }  && cd student || cd / && cd ;

...is equivalent to...

pwd; { { { (ls) || { cd .. && ls student/; }; } && cd student; } || cd /; } && cd ; }

...breaking that down graphically:

pwd; {                                      # 1
       {                                    # 2
         { (ls) ||                          # 3
                   { cd .. &&               # 4
                              ls student/;  # 5
                   };                       # 6
         } && cd student;                   # 7
       } || cd /;                           # 8
     } && cd ;                              # 9

1. pwd happens unconditionally
2. (Grouping only)
3. ls happens (in a subshell) unconditionally.
4. cd .. happens if (3) failed.
5. ls student/ happens if (3) failed and (4) succeeded
6. (Grouping only)
7. cd student happens if either (3) succeeded or both (4) and (5) succeeded.
8. cd / happens if either [both (3) and one of (4) or (5) failed], or [(7) failed].
9. cd happens if (7) occurred and succeeded, or (8) occurred and succeeded.

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Using explicit grouping operators is wise to avoid confusing yourself. Avoiding writing code as hard to read as this is even wiser.

(ls) is executed in a subshell, because the commands are separated with ";"

(ls) is executed in a subshell because of the parentheses. Parentheses introduce subshells.

But I have no idea why other commands are executed?

Unlike other programming languages you might be familiar with, bash does not give && higher precedence than ||. They have equal precedence and are evaluated from left to right.

If you had a || b && c, in other languages this would be read as a || { b && c; }. If a were true then neither b nor c would be evaluated.

In bash, though, it's parsed as { a || b; } && c (strict left-to-right precedence), so when a is true b is skipped but c is still evaluated.