Definition of "well defined" in mathematics
The term well-defined (as oppsed to simply defined) is typically used when a definition seemingly depends on a choice, but in the end does not.
In most (but not all) cases, this applies to the definition of a function $f\colon A\to B$ in terms of two given functions $g\colon C\to A$ and $h\colon C\to B$: For $a\in A$ we want to define $f(a)$ by first picking an element $c\in C$ with $g(c)=a$ and then let $f(a)=h(c)$. Two problems arise with this: First of all, we must make sure that for each $a\in A$ there exists $c\in C$ with $g(c)=a$, in other words: $g$ must be surjective. But we also must make sure that the choice of $c$ is irrelevant, that is: Whenever $g(c)=g(c')$ it must also be true that $h(c)=h(c')$. Only if $g,h$ fulfil these conditions the above construction will actually define a function $f\colon A\to B$. To repeat: After this, $f$ is in fact defined. The definition itself does not become a "better" definition by saying that $f$ is well-defined. Instead, saying that $f$ is well-defined just states the (hopefully provable) fact that the conditions described above hold for $g,h$, and so we really have given a definition of $f$ this way. If the conditions don't hold, $f$ is not somehow "less well defined", it is not defined at all.
There is an additional, very useful notion of well-definedness, that was not written (so far) in the other answers, and it is the notion of well-definedness in an equivalence class/quotient space.
Take an equivalence relation $E$ on a set $X$. We can then form the quotient $X/E$ (set of all equivalence classes). Take another set $Y$, and a function $f:X\to Y$. If $f(x)=f(y)$ whenever $x$ and $y$ belong to the same equivalence class, then we say that $f$ is well-defined on $X/E$, which intuitively means that it depends only on the class.
More rigorously, what happens is that in this case we can ("well") define a new function $f':X/E\to Y$, as $f'([x])=f(x)$.
As an example, take as $X$ the set of all convex polygons, and take as $E$ "having the same number of edges". The number of diagonals only depends on the number of edges, and so it is a well-defined function on $X/E$.
The statement '' well defined'' is used in many different contexts and, generally, it means that something is defined in a way that correspond to some given ''definition'' in the specific context.
As a simple example: if I say:
given the function $f(x)=\sqrt{x}=y$ such that $y^2=x$
this function is not well defined. A function is well defined only if we specify the domain and the codomain, and iff to any element in the domain correspons only one element in the codomain.
So, $f(x)=\sqrt{x}$ is ''well defined'' if we specify, as an example, $f : [0,+\infty) \to \mathbb{R}$ (because in $\mathbb{R}$ the symbol $\sqrt{x}$ is, by definition the positive square root) , but, in the case $ f:\mathbb{R}\to \mathbb{C}$ it is not well defined since it can have two values for the same $x$, and becomes ''well defined'' only if we have some rule for chose one of these values ( e.g. the principal square root).
As another example: if I say:
consider the vector space $\mathbb{R}^n$
this is not a well defined space, if I not know what is the field over which the vector space is given. $\mathbb{R}^n$ over the field of reals is a vectot space of dimension $n$, but over the field of rational numbers it is a vector space of dimension uncountably infinite.
Also for sets the definition can gives some problems, and we can have sets that are not well defined if we does not specify the context. As an example consider the set
$D=\{x \in \mathbb{R}: x \mbox{ is a definable number}\}$
Since the concept of ''definable real number'' can be different in different models of $\mathbb{R}$, this set is well defined only if we specify what is the model we are using ( see: Definable real numbers)
Obviously, in many situation, the context is such that it is not necessary to specify all these aspect of the definition, and it is sufficient to say that the thing we are defining is '' well defined'' in such a context.