问题

第一件事是你没有为你的参数使用泛型类型，这将导致typescript永远不会根据你的输入推断出正确的类型（你可以想象泛型类型是参数，tsc需要它根据你的输入计算结果）。
简而言之
const handleReaction = (arg: Argument): Argument extends { type: "status" } ? Status : GoalActivity => { // ... }
将始终返回Status | GoalActivity作为返回类型。

溶液

当然，这里你必须使用泛型类型作为你的参数，我将用行内解释来拆分你的代码：

type Status = 'statusType'
type GoalActivity = 'goalActivityType'

type StatusObj = { type: 'status'; status: Status | null };
type GoalActivityObj = { type: 'goalActivity'; goalActivity: GoalActivity | null }

type Argument = StatusObj | GoalActivityObj;

// Define returned type based on a input argument `T`
type ReturnType<T> = T extends StatusObj ? Status : GoalActivity;

// Generic type should be used here
const handleReaction = <T extends Argument>(arg: T): ReturnType<T> => {
    if (arg.type === 'status') {

        // Q: Why do we have to cast here?
        // A: Any returned type can't assign to statement of `type ReturnType<T> ...`
        // but luckily `tsc` still allows us to cast back since they are all string literal
        return 'statusType' as ReturnType<T>;

    } else {
        return 'goalActivityType' as ReturnType<T>;
    }
}

你的参数对象应该有自己的类型。一旦它们有了类型，你就可以像这样使用函数重载了：

type Status = { type: 'status', status: 'statusValue' };
type GoalActivity = { type: 'goalActivity', goalActivity: 'goalValue' };

function handleReaction(arg: Status): Status['status']
function handleReaction(arg: GoalActivity): GoalActivity['goalActivity']
function handleReaction(arg: Status | GoalActivity) { 
    if ('status' in arg) {
        return arg.status;
    } else {
        return arg.goalActivity;
    }
}

const a = handleReaction({ type: 'status', status: 'statusValue' })
//    ^? 'statusValue'
const b = handleReaction({ type: 'goalActivity', goalActivity: 'goalValue' })
//    ^? 'goalValue'

Playground