Why are vector valued functions 'well-defined' when multivalued functions aren't?

What you are getting confused by most probably is the fact that a vector seems to be multiple elements, but really it's not, it's just a single element which (often) has multiple parts.

You can even have so called set-valued functions. An easy one to think of is the function which takes a positive integer to the set of its prime divisors. Let's call this function $\pi$.

The important thing in both these cases is that a given input only has one output. So in the case of a vector-valued function the output is the full vector (all of its parts). In the case of the set valued function for prime divisors you have to get the set of all the prime divisors. So for example $\pi(15)=\{3,5\}$ you can't have $\pi(15)=\{3\}$ or $\pi(15)=\{5\}$. There is only the one set containing all the divisors. It's the same situation with vector-valued functions. There is only one vector for any input. The function can't return just part of a vector. Thus we still have only one output for every input--it's just a bigger output.

As to why we don't define inverses as vector valued, that has a very easy reason. They wouldn't be inverses anymore. The definition of an inverse to a function $f$ is a function $g$ for which $g(f(x))=x$ if the value of $g$ is a vector containing $x$ that doesn't do what we want. It doesn't tell us the value of the original input to function $f$.

Your intuition seems fine; the issue seems primarily linguistic.

Let's start with the usual (mildly pedantic) definition of a "function" in terms of sets: If $X$ and $Y$ are non-empty sets, their Cartesian product $X \times Y$ is the set whose elements are ordered pairs $(x, y)$ with $x \in X$ and $y \in Y$. A function $f:X \to Y$ is a subset of $X \times Y$ satisfying the following property:

For every $x$ in $X$, there exists a unique $y$ in $Y$ such that $(x, y) \in f$.

The conceptual content is that each input $x$ uniquely determines the output $y = f(x)$. Philosophically, a function can be viewed as a mathematical formalization of determinism: If initial conditions (i.e., $x$) are known, then a subsequent state (i.e., $y$) is uniquely determined.

Note that to avoid verbal ambiguity, mathematicians always (at least implicitly) specify the domain $X$ (the set of inputs) and the target $Y$ (the set of potential outputs) when speaking of a function. (Non-mathematicians can often afford to be more lax, writing down formulas or the equivalent and speaking of "the corresponding function".)

It's certainly useful in some circumstances to allow "multi-valued functions". A multi-valued function $F:X \to Y$ might be defined as a subset of $X \times Y$ satisfying the property:

For every $x$ in $X$, there exists a $y$ in $Y$ such that $(x, y) \in F$.

Just as in computer science, a multi-valued function $F:X \to Y$ may be "interpreted as" (or, more precisely, "converted to") a single-valued function by allowing the function to return a "compound data structure" rather than an element of $Y$.

To convert a multi-valued function into a single-valued function, we might view $F$ as taking values in the power set $\mathscr{P}(Y)$, whose elements are subsets of $Y$.

For example, if $X = [0, \infty)$ and $Y = (-\infty, \infty)$ denote intervals of real numbers, the set $$ F = \{(x, y)\text{ in } X \times Y: y^{2} = x\} $$ would be the (generally double-valued, i.e., not well-defined) "square root function", $F(x) = \pm\sqrt{x}$.

The "value" of $F$ at a positive real number $x$ could, however, be viewed as the set (a.k.a., unordered pair) $\{\sqrt{x}, -\sqrt{x}\}$. Doing so would amount to defining a single-valued (i.e., well-defined) function $F_{1}:[0, \infty) \to \mathscr{P}\bigl((-\infty, \infty)\bigr)$ in "the obvious way": If $0 \leq x$, then $y \in F_{1}(x)$ if and only if $y^{2} = x$, if and only if $y = \sqrt{x}$ or $y = -\sqrt{x}$. As a formula, $$ F_{1}(x) = \{-\sqrt{x}, \sqrt{x}\}. $$

Alternatively, the "value" of $F$ could be viewed as an ordered pair: $(-\sqrt{x}, \sqrt{x})$. Doing so would amount to defining a single-valued function $F_{2}:[0, \infty) \to (-\infty, \infty) \times (-\infty, \infty)$. As a formula, $$ F_{2}(x) = (-\sqrt{x}, \sqrt{x}). $$

(One can imagine returning even more exotic objects, such as probability measures on a set of elements. The point is, there's not a unique way of making a multi-valued function into a single-valued function.)

Again, the primary issue seems linguistic. The single relation $F$ above may be viewed as "double-valued" in the real numbers (not well-defined); as single-valued in sets of reals; or as single-valued in ordered pairs of reals. But there's no terminological ambiguity, because a mathematician would view these as three distinct entities.

Your terminology is confusing because there is such a thing as a vector valued function, that's a legitimate (and well defined) function into something like $\mathbb R\times\mathbb R$. What you mean is "why can't a function take two different values on the same input?" Such a thing does exist, it's called a relation. But a relation is not a function unless it has an extra property. There's no harm in working with relations, they just don't satisfy all the nice properties functions do, so you really need functions and not just relations to make headway.