[typescript-fetch] Remove returns in oneOf serialization code which short circuit additional type checks

[cgual206]

Fixes #21811

The JSON serialization/deserialization code generated for oneOf schemas which do not use a discriminator follows a pattern of validating & serializing first the model schemas, then arrays of non-primitive objects, then primitive objects (including arrays).

The area of the template which validates & serializes / deserializes arrays of non-primitive objects includes a return value; or return json; statement if it does not find a matching object. For example:

    if (Array.isArray(value)) {
        if (value.every(item => typeof item === 'object')) {
            if (value.every(item => instanceOfTestA(item))) {
                return value.map(value => TestAToJSON(value as TestA));
            }
            if (value.every(item => instanceOfTestB(item))) {
                return value.map(value => TestBToJSON(value as TestB));
            }
        }
        return value;
    }

This return short circuits further validation / serialization that should happen if the schema is an array of a primitive type.

Additionally there are a couple of other fixes for:

• Code which is deserializing arrays of Dates is using the entire json string instead of deserializing each item in the array.
• Variable name duplication in the lambdas for some calls to .map (e.g. value.map(value => value.toISOString()))
• Not casting array items as object when calling instanceOf methods
• Several paces where the code which has unbalanced parenthesis

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[WIStudent]

Instead of using item as object we could also tell typescript that json is object[] if the condition that checks that every item is object is true:

if (json.every((item): item is object => typeof item === 'object')) {
    if (json.every(item => instanceOfTestA(item))) {
      // ...
    }
}

[WIStudent]

Another thing: The check, that every item in the array is an object should also check that the item is not null, because typeof null also returns 'object'.

if (json.every((item): item is object => item !== null && typeof item === 'object')) {
// ...
}

Otherwise instanceOfTestA will be called with null if json contains [null], and that will cause a TypeError because the in operator cannot be used with null.

export function instanceOfTestA(value: object): value is TestA {
    if (!('foo' in value) || value['foo'] === undefined) return false;
    return true;
}