I just want to give a refinement of other answers so far, as well as a different point of view (namely, that of a person who knows little about set theory but who also encounters these kinds of issues).

As others have mentioned, the root cause of the problem is that there are big logical problems with considering "the set of all sets". Unless you would like to learn more about set theory, you needn't concern yourself with what these problems are (though Russel's paradox is fairly elementary and kind of fun). It is just one of those facts of life that non-set theorists learn to live with and that set theorists learn to love. The non-existence of the set of all sets forces us to abandon other putative sets, such as "the set of all groups", "the set of all vector spaces", "the set of all manifolds", etc. For example, it is possible to equip any set with the structure of a group and so if we were able to build the set of all groups then we would necessarily have also build the set of all sets. This is almost always what people mean when they claim that a certain construction is "too big to be a set" - the construction invokes a sloppy use of set theory language that taken literally accidentally constructs the set of all sets as a byproduct. In your case, the existence of the set of all bilinear maps on $V \times W$ constructs as a byproduct the set of all vector spaces over $R$ (every bilinear map has to have a target), and if there were a set of all vector spaces over $R$ then there would be a set of all sets.

This is probably not the last time you will encounter this sort of issue. In basically every case, however, there is a trick that swoops in and saves the day. Generally the idea is to observe that you don't actually need all of the flexibility that you tried to give yourself by constructing a non-set, and that it is enough to consider a simpler object (in your case the set of all bilinear maps from $V \times W$ to $\mathbb{R}$) which is small enough to be a set but big enough to have the property that you want (in your case you want it to function as a sort of universal bilinear pairing between $V$ and $W$).

Ultimately I regard these sorts of concerns as analogous to the "end user agreements" that you have to certify you've read whenever you install a Microsoft product or sign up for a gmail account. I'm sure all that fine print is important, but I feel like I would have to become a lawyer to understand it all. And just as in that case, you don't have to be a set theorist to understand how to resolve these sorts of issues most of the time - usually it just requires you to capture the flexibility present in what you are already working on.

Just recently I was reading about an object which was given as the quotient by a certain equivalence relation of the set of all pairs $(T, H)$ where $H$ is a Hilbert space and $T$ is a certain kind of operator on $H$. The book pointed out that one cannot consider the set of all Hilbert spaces, but that the set theoretic difficulties can be resolved by proving that every pair $(T', H')$ is equivalent to a pair $(T, H)$ on a fixed Hilbert space $H$. So the problem was avoided by exploiting some inherent flexibility in the equivalence relation under consideration. This sort of behavior is quite typical.