R, 96 65 72 bytes

Or R>=4.1, 65 bytes by replacing the word function with \.

-many bytes by changing I/O to vector of char codes.
-another some bytes thanks to @Dominic van Essen.
+7 bytes costed converting from not-so-self-contained function to a program.

i=0;s=scan();f=function(a=s[i<<-i+1])"if"(a<58,a,c(40,f(),a,f(),41));f()

Try it online!

Port of @loopy walt's answer.

Explanation

Initiates global counter i to 0 and takes input with scan.
The recursive function f takes one character at a time (a) incrementing the global counter each time. If the character char code is less then 58, then it's a digit and needs to be returned as is. Otherwise, we surround the character with braces (char codes 40,41) and recursive calls to our function.

Solution shorter for R>=4.1:

R, 74 bytes

Or R>=4.1, 60 bytes by replacing two function occurrences with \s.

function(s,i=0,f=function(a=s[i<<-i+1])"if"(a<58,a,c(40,f(),a,f(),41)))f()

Try it online!

Same as above, but here i is "local" for the main function, but "global" for the recursive one.

R, 96 65 72 69 bytes

Or R>=4.1, 62 bytes by replacing the word function with \.

-many bytes by changing I/O to vector of char codes.
-another some bytes thanks to @Dominic van Essen.
+7 bytes costed converting from not-so-self-contained function to a program.
-3 bytes thanks to @Dominic van Essen - converting back to one (reusable) function.

`/`=function(s,t=0)"if"((a=s[i<<-t*i+1])<58,a,c(40,s/1,a,s/1,41));i=9

Try it online!

Port of @loopy walt's answer.

Explanation

A function takes string s and t defaulting to 0 (for resetting global variable).
The recursive function (named / overriding division) takes one character at a time (a) incrementing the global counter each time. If the character char code is less then 58, then it's a digit and needs to be returned as is. Otherwise, we surround the character with braces (char codes 40,41) and recursive calls to our function (with t=1 to keep i as is between calls).
The code at the end initiates global counter i to a digit (9 chosen for demonstration purposes).

Solution shorter for R>=4.1:

R, 74 bytes

Or R>=4.1, 60 bytes by replacing two function occurrences with \s.

function(s,i=0,f=function(a=s[i<<-i+1])"if"(a<58,a,c(40,f(),a,f(),41)))f()

Try it online!

Same as above, but here i is "local" for the main function, but "global" for the recursive one.

R, 96 65 72 bytes

Or R>=4.1, 65 bytes by replacing the word function with \.

-many bytes by changing I/O to vector of char codes.
-another some bytes thanks to @Dominic van Essen.
+7 bytes costed converting from not-so-self-contained function to a program.

i=0;s=scan();f=function(a=s[i<<-i+1])"if"(a<58,a,c(40,f(),a,f(),41));f()

Try it online!

Port of @loopy walt's answer.

Explanation

Initiates global counter i to 0 and takes input with scan.
The recursive function f takes one character at a time (a) incrementing the global counter each time. If the character char code is less then 58, then it's a digit and needs to be returned as is. Otherwise, we surround the character with braces (char codes 40,41) and recursive calls to our function.

R, 96 65 72 bytes

Or R>=4.1, 65 bytes by replacing the word function with \.

-many bytes by changing I/O to vector of char codes.
-another some bytes thanks to @Dominic van Essen.
+7 bytes costed converting from not-so-self-contained function to a program.

i=0;s=scan();f=function(a=s[i<<-i+1])"if"(a<58,a,c(40,f(),a,f(),41));f()

Try it online!

Port of @loopy walt's answer.

Explanation

Initiates global counter i to 0 and takes input with scan.
The recursive function f takes one character at a time (a) incrementing the global counter each time. If the character char code is less then 58, then it's a digit and needs to be returned as is. Otherwise, we surround the character with braces (char codes 40,41) and recursive calls to our function.

Solution shorter for R>=4.1:

R, 74 bytes

Or R>=4.1, 60 bytes by replacing two function occurrences with \s.

function(s,i=0,f=function(a=s[i<<-i+1])"if"(a<58,a,c(40,f(),a,f(),41)))f()

Try it online!

Same as above, but here i is "local" for the main function, but "global" for the recursive one.

R, 96 65 bytes

Or R>=4.1, 58 bytes by replacing the word function with \.

-many bytes by changing I/O to vector of char codes.
-another some bytes thanks to @Dominic van Essen.

f={i=0;function(s,a=s[i<<-i+1])"if"(a<58,a,c(40,f(s),a,f(s),41))}

Try it online!

Port of @loopy walt's answer.

Explanation

Creates local environment with {} and initiates counter i to 0.
The environment returns a function which is assigned to f.
The recursive function inside takes one character at a time (a) incrementing the global counter each time. If the character char code is less then 58, then it's a digit and needs to be returned as is. Otherwise, we surround the character with braces (char codes 40,41) and recursive calls to our function.

R, 96 65 72 bytes

Or R>=4.1, 65 bytes by replacing the word function with \.

-many bytes by changing I/O to vector of char codes.
-another some bytes thanks to @Dominic van Essen.
+7 bytes costed converting from not-so-self-contained function to a program.

i=0;s=scan();f=function(a=s[i<<-i+1])"if"(a<58,a,c(40,f(),a,f(),41));f()

Try it online!

Port of @loopy walt's answer.

Explanation

Initiates global counter i to 0 and takes input with scan.
The recursive function f takes one character at a time (a) incrementing the global counter each time. If the character char code is less then 58, then it's a digit and needs to be returned as is. Otherwise, we surround the character with braces (char codes 40,41) and recursive calls to our function.