Understand two examples using indirect expansion for variable expansion in arithmetic expressions

Parameter expansion isn’t the same as arithmetic evaluation.

In an arithmetic expression, a text string such as a is interpreted as a variable name, and the value of that variable is processed as an arithmetic expression itself, and the result used in the containing arithmetic expression. Thus

$((a * 3))

causes the value of the variable a, 1+1, to be evaluated as an arithmetic expression, yielding 2, and the whole expression becomes $((2 * 3)), i.e. 6.

In the same context, $a is replaced with the value of a as-is, in an earlier phase (this is parameter expansion); so

$(($a * 3))

becomes $((1+1 * 3)), which is 4 following the usual precedence rules.

In the second example, $((++a)) is processed before $b (processing takes place from left to right). The arithmetic expression causes a to be evaluated as an arithmetic expression itself, as above; the value of a is a=b++, which when evaluated, assigns the value of b (itself evaluated as an arithmetic expression) to a, then increments b. Thus after evaluating a, b is 2 and a is 1. Evaluating $((++a)) increments a and returns its value, which is now 2. $b is just replaced with the value of b, 2 as well. After the quoted argument to echo is processed, the echo is run with 2 2 as its sole argument, producing the output you saw.