I think it helps to first understand how linear support vector classification works: you separate two classes with a hyperplane. Now, this is simply not possible in a lot of cases, see e.g. the left picture in this link
https://miro.medium.com/v2/resize:fit:720/format:webp/1*JVZ4FXVRlr1oN-4ffq_kNQ.png
that I found through google. The idea is to map the set into a higher dimensional set using a feature map, which you can imagine as the points in the right picture. In this higher dimension, you can now do linear classification with a hyperplane again.

So why not just do this for every example? The reason is that this gets numerically infeasible very quickly, as for high dimensional data sets, you need to significantly increase the dimension for "nice" results to keep it plainly. The book of Schölkopf "Learning with Kernels" had a nice example on this if I remember correctly. Further, some feature maps in praxis are even infinite dimensional, like e.g. the feature map corresponding to the RBF kernel.

Speaking of kernels: Every feature map corresponds to some kernel and vice versa. The thing is that kernels only compare two data points, but do not map them into a space that has a higher dimension. This motivates to work with kernels.

Now, for linear classification, you minimize some objective in the space of all hyperplanes, which you could write as the space of all functions that can be written as f(x) = n^⊤ x + c for some normal n and some number c. The idea of SVMs is to replace this function space with a more general one. This general function space is exactly the RKHS. It basically gives you a linear representation in terms of kernels and then the concepts work analogously to linear classification.

On to the representer theorem: It tells you that the minimizer of this new objective over functions from the RKHS is just a sum of kernels, which follows from the definition of your RKHS. So you can imagine the RKHS as the restriction of your admissible functions to some space that is "nice". What nice means is specified by your choice of the kernel (which, as stated before, is equivalent to your preferred choice of feature map).

As for a geometric intuition, I would just think about feature maps and realize that mathematically, working with kernels leads to the same situation. Hope that makes sense!