Do we need to check all 10 axioms to verify that some set, call it $V$, is a vector space?

Take for example the subset $V = \mathbb{R}_{\geq 0} \subset \mathbb{R}$ with the usual addition and define the scalar multiplication via $k \cdot x = \vert k \vert x$. Then 1) and 6) hold, but 5) does for example not hold as we do not have negative real numbers in $V$.

Part I. I'm going to give a meta-answer here: most definitions in elementary mathematics (let's say "up to junior year in college") are pretty well established and tested. There are some bad books/papers out there, but if four different linear algebra books define "vector space" the same way, you can bet that there's a reason.

Now what about redundancy? Do you really need all those properties? You bet. If there were a way to prove, say, the 7th property from the others, every book would have only 9 rules, and then have a little theorem saying that a system that satisfies those 9 rules also satisfies this property (the 7th one from the original set).

Why is this? Because mathematicians, just like you, don't like having to confirm 15 things when confirming 10 is enough. So they do their very best to make the definitions minimal. As you've seen from the examples here, if you remove one of the properties, someone can usually come up a a system that has the other 9, but not the one you removed, i.e., they can show that the removed property is not a consequence of the other 9, so it's really necessary.

Part II. Do mathematicians really go through all those steps every time they encounter something and want to prove it's a vector space? Well, yes and no. After you've done a few of these proofs, it gets so easy that you can just say to yourself, "yep...that's gonna work out..." and not write down the details. But if someone claims that something unusual is a vector space, they'll usually sketch out the reasons for at least a bunch of the properties. And perhaps most important, folks tend to prove things like "If we have a set $S$ and consider the collection $Q$ of all formal sums of elements of $S$ with coefficients in $F_2$, the field of two elements, then $Q$ has a natural vector space structure." Then one can apply this to any set $S$, while only doing the "is a vector space" proof once.

Part III. Your particular text's definition of a vector space is kinda...weak, at least if you transcribed it faithfully. So my claim in part 1 is a little shaky. In particular, condition 5 should read $u + (-u) = 0$, rather than $u + (-v) = 0$, but I assume that this was just a miscopying on your part. The real version of condition $5$ is that there should be, for each vector $u$, exactly one other vector $v$ with the property that $u + v = 0$; that vector $v$ is denoted $-u$. As written, there's an assumption that your set comes with a negation operation, which wasn't mentioned anywhere. In fact, the whole definition's a bit of a mess, because the author didn't want to say that a vector space consisted of a set together with an operation (called vector addition), a field, and another operation (call scalar multiplication) from pairs $(c, v)$ in the field and the set to items in the set, and that these two operations had to have certain properties. Perhaps the author didn't want to define "operation" carefully, or maybe the author was just lazy. I can't say, but a text that defined vector space this way wouldn't be one of my treasured friends...

Consider the set $S$ of $2 \times 2$ real matrices and endow it with a $+$ law which is in fact the product of matrices. With that axiom 1 is satisfied.

And say the multiplication by a real is the usual one. Therefore axiom 6 is verified.

We know that product of matrices in not commutative. So axiom 2 is not fulfilled.

You can look at this post to dig further into the topic.