What do mathematicians mean by "equipped"?

The word "equipped" keeps notational pandemonium from breaking loose. For instance, if you were to be a bit more formal, you'd say

A Hilbert space is a pair $(V, \left<\cdot,\cdot \right>)$, where $V$ is a vector space and $\left<\cdot,\cdot \right>\colon V\times V \to \mathbb{C}$ is an inner product. Additionally, all Cauchy sequences in $V$ are convergent in the norm induced by the inner product to an element in $V$.

But most of the time, there's no reason to disambiguate between the vector space and the inner product (who puts a different inner product on the set $L^2[0,1]$?), so we refrain from defining these "pairs", and simply "equip" our space.

Generally, when we use the word "equipped", we mean not that it is possible to do an operation, but we have one in mind. That is on a vector space like $\mathbb R^2$, we could think of various inner products that would suffice. For instance the following two operations are both inner products: $$(x_1,x_2)\cdot(y_1,y_2)=x_1y_1+x_2y_2$$ $$(x_1,x_2)*(y_1,y_2)=(x_1+x_2)(y_1+y_2)+x_2y_2$$ but they are different. That means that, when we choose an inner product, we are adding additional structure to $\mathbb R^2$ - just in a plain old vector space, we have no notion of orthogonal, and we can choose an appropriate inner product to make any pair of vectors orthogonal. For instance, in the above example, under the inner product $\cdot$, we have that $(0,1)$ and $(1,0)$ are orthogonal, but under the second inner product $*$ they have product $1$ and are not orthogonal (however, the vectors $(1,-1)$ and $(1,0)$ are, which is not true of the first metric). Similarly, notions like a unit vector cannot be defined in a vector space, as different inner products will give different answers. Thus, we must equip the vector space with some particular inner product before we can sensibly talk about an inner product - which means that we're keeping the old structure (of a vector space), but adding some new structure on top of it.

A notation, which is fairly prevalent in the field of general abstract nonsense, is to write something like "a set $G$ equipped with a binary operation $\cdot:G\times G\rightarrow G$" would be to refer to it as a tuple $(G,\cdot)$ - that is, we take a group (or magma) not to be just the set on which we have some external operation, but to take both set and operation into a single structure, as one is meaningless without the other.

I think the best way of explaining what it means to be equipped with something is by showing how the term is used. Mathematics has a very precise formalism, but in its pure form it is mostly not appropriate for daily use because then even the simplest problems would become very hard. Equipping a mathematical object with additional structures is one of the tricks for allowing us to use shortcuts we are familiar with from daily life without giving up too much precision.

E.g., we could define a mounted policeman as a policeman P equipped with a horse H. Formally this means that a policeman is an ordered pair (P,H) consisting of a policeman and a horse. In reality there are some unstated assumptions that there is actually a certain connection between the policeman and the horse. (The policeman has to be sitting on it. Or at least he must have been assigned the horse for today's shift.) In mathematics, these tend to be more obvious and less ambiguous.

Technically we would have to distinguish strictly between the policeman P and the mounted policeman M = (P,H) of which he is the first constituent. E.g., if the policeman has a beard B(P), it just means that the first element of the mounted policeman (the policeman, not the horse or the pair) has a beard. Of course that's not how we say it in practice, and that's precisely the point. Even pure mathematicians will just write B(M) in most situations (rather than something like B(P(M))) because ultimately, we think of a mounted policeman as primarily a policeman. This becomes clear if you think of the tail of the mounted policeman. There is no such thing, even though the horse probably has one! This is why it's absolutely not the same to equip a policeman with a horse, or a horse with a policeman.

In fact, if we think of a policeman as a man who is equipped with a certain job, then saying that he has a beard really means that the man part m of the policeman has a beard. And if a man is a person equipped with masculinity and maturity, then for a man to have a beard really means that the person has a beard.

We can also think of a mounted policeman as a mounted policeperson who happens to be male, or as a mounted man who happens to have a certain job. Technically all these interpretations give rise to different definitions of a mounted policeman. But these differences are harmless and in most contexts (even most mathematical contexts) can be ignored: It is harmless to identify the different definitions of a mounted policeman.

In formulas, if m is the man, J is his job and H is his horse, then it doesn't matter in practice whether we think of the policeman as (m,J) and therefore of the mounted policeman as ((m,J),H), or if we think of the mounted man as (m,H) and therefore of the mounted policeman as ((m,H),J). So long as we remember where we have put the horse and where we have put the job, it's always clear how to translate from one representation to the other. Formally, a man equipped with a police job, and then equipped with a horse, is different from a man equipped with a horse and then equipped with a police job. But in practice we identify these two kinds of objects. I.e. we consider two mounted policemen to be the same if the underlying men are the same, the equipped horses are the same and the equipped jobs are the same.

The last point is different from natural language use, where we would consider M1 = (Chief Inspector Derrick, Rosie) and M2 = (Chief Inspector Derrick, Jack) to be the same mounted policeman. In mathematics we would instead be more precise and say that M1 and M2 are the same as (unequipped) policemen.