JavaScript (ES6), 67 bytes

f=([c,...a])=>c?c+f([...a.splice(0,a.indexOf(c)).reverse(),...a]):a

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JavaScript (ES6), 67 bytes

f=([c,...a])=>c?c+f([...a.splice(0,a.indexOf(c)).reverse(),...a]):a

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How?

a.splice(0, n) removes and returns the first \$n\$ entries in a[] if \$n\ge1\$ or removes nothing and returns an empty array if \$n<1\$. If c is not found in a[], a.indexOf(c) returns \$-1\$ and we just pass a[] unchanged to the next iteration.

JavaScript (ES6), 84 bytes

f=([c,...a])=>c?c+f(~(i=a.indexOf(c))?[...a.slice(0,i).reverse(),...a.slice(i)]:a):a

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JavaScript (ES6), 67 bytes

f=([c,...a])=>c?c+f([...a.splice(0,a.indexOf(c)).reverse(),...a]):a

Try it online!