What's the explanation for Exercise 1.6 in SICP?

new-if is a function. When a function is called, what's the first thing that Scheme does with the argument list? It evaluates all the arguments.

new-if is a procedure, and Scheme uses applicative-order evaluation (1.1.5), so even before new-if is actually performed, it has to evaluate all the arguments first, which are guess and (sqrt-iter (improve guess x) x). You can see that the latter argument is a recursion, which calls a new new-if procedure, this is how the infinite loop occurs.

The ordinary if need not evaluate its arguments first, just go along the way, this is the difference between if and new-if. :)

First of all you have to understand the difference between applicative order evaluation and normal order. Lisp uses applicative order, but conditional expressions are evaluated not like normal functions (sicp chapter 1.1.6):

(if <predicate> <consequent> <alternative>)

To evaluate an if expression, the interpreter starts by evaluating the <predicate> part of the expression. If the <predicate> evaluates to a true value, the interpreter then evaluates the <consequent> and returns its value. Otherwise it evaluates the <alternative> and returns its value.