TypeScript: How to get nested functions to union their types through a chain

You are trying to change the generics of withZoo(..) based upon the expressions around it which is not possible. You could create a single generic that takes in these middleware-like callbacks then use the data from those callbacks to type a single callback function like the following:

export type SomeHandler<T> = (args: T) => void;

type WithFooArgs = { foo: string }
const withFoo = <T,>(handler: SomeHandler<T & WithFooArgs>) => (args: T & WithFooArgs) => handler(args);

type WithBarArgs = { bar: string }
const withBar = <T,>(handler: SomeHandler<T & WithBarArgs>) => (args: T & WithBarArgs) => handler(args);

type WithZooArgs = { zoo: string }
const withZoo = <T,>(handler: SomeHandler<T & WithZooArgs>) => (args: T & WithZooArgs) => handler(args);

// https://stackoverflow.com/a/50375286/10873797
type UnionToIntersection<U> = 
  (U extends any ? (k: U)=>void : never) extends ((k: infer I)=>void) ? I : never;

function createHandler<Handler extends SomeHandler<any>>(handlers: Handler[], cb: (args: UnionToIntersection<Parameters<Parameters<Handler>[0]>[0]>) => void) {
    return (args: any) => {
        return cb(handlers.reduce((acc, cur) => cur(acc), args));
    };
}

export default createHandler([withFoo, withBar, withZoo], (args) => {
    console.log(args); // WithFooArgs & WithBarArgs & WithZooArgs
});

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