Homotopy theory of function spaces is a healthy subfield of homotopy theory. See e.g. recent Oberwolfach report from a meeting on the subject.
If you share what $X,Y$ you are interested in, I may be able to say more, even though I am by no means an expert.
EDIT: In response to your edit, suppose $\Sigma$ is a compact Riemann surface, $G$ is a compact Lie group, and let's study $k$th homotopy group of $C^0(\Sigma, G)$. Actually, most of the discussion below applies to general $X,Y$ that are, say, compact manifolds or even finite CW-complexes, but let's stick to the case of interest.
There is an well-known homeomorphism $C^0(S^k,C^0(\Sigma, G))\cong C^0(S^k\times\Sigma, G)$ given by the adjoint, and since any homeomorphism preserves path-components, we can identify $\pi_k(C^0(\Sigma, G))$ with $[S^k\times \Sigma, G]$, the set of homotopy classes of maps from $S^k\times \Sigma$ to $G$. Smashing $S^k\vee\Sigma$ inside $S^k\times\Sigma$ to a point gives $S^k\wedge\Sigma$, which is the $k$-fold suspension $S^k\Sigma$ of $\Sigma$. The cofiber sequence of the quotient map $S^k\times\Sigma\to S^k\wedge\Sigma$ is an exact sequence, namely,
$$[S^k\Sigma, G]\to [S^k\times\Sigma, G]\to [S^k\vee \Sigma, G]=\pi_k(G)+[\Sigma, G].$$
Let's suppose that $G$ is simply-connected, so that $[\Sigma, G]$ is a point (as any Lie group has trivial $\pi_2(G)$ and $\Sigma$ is $2$-dimensional). Then the above cofiber sequence is an exact sequence of groups (not just sets). In trying to compute $[S^k\Sigma, G]$ it helps to recall that $G$ is rationally homotopy equivalent to the product of odd-dimensional spheres, so rationally $[S^k\Sigma, G]$ is the product of $[S^k\Sigma, S^m]$'s where $m$ is odd, and of course $[S^k\Sigma, S^m]=[\Sigma, \Omega^kS^m]$. This should allow you to do some computations.
Finally, I wish to comment that the inclusion $C^\infty(X,Y)\to C^0(X,Y)$ is a weak homotopy equivalence i.e. it induces an isomorphism of homotopy groups as any map of a sphere/disk is homotopic to a nearby smooth map. A result of Milnor says that $C^0(X,Y)$ is homotopy equivalent to a CW-complex, provided $X$ is a finite CW-complex (if memory serves me). I am not sure why the same is true for $C^\infty(X,Y)$ so at the moment I do not know if the above inclusion is a homotopy equivalence.
I think there are a few things to untangle here.
First, as concerns your highlighted question, it seems that you've answered it yourself: outside the compact Hausdorff case (where the uniform structure is completely equivalent to the topology), it's unreasonable to think that the associated condensed set "knows" the uniform structure. It really only knows the topology.
Second, you are correct in your unpacking of the definition of "solid" as it applies to (let's say metrizable) toplogical abelian groups $M$. If $\underline{M}$ is solid, then for any nullsequence $(m_n)$ in $M$ and any sequence of integers $(a_n)$ the sum $\sum a_nm_n$ must converge in $M$. I agree that this feels a lot saying that $M$ is nonarchimedean and complete. However, though I gather for you "complete" means "Cauchy-complete" as usual, I'm not sure what general definition of a topological abelian group $M$ being "nonarchimedean" you're thinking of here to say that you think it is actually equivalent.
Here's something I find helpful to keep in mind. There are several differences between the notion of "solid" and the notion of "complete", beyond the fact that solid enforces some kind of non-archimedeanness. (Indeed, the same remarks apply in the liquid setting, which does not enforce nonarchimedeanness.)
First, while both the definition of "complete" and the definition of "solid" are of the form "for every [...] there exists a unique {...}", the differences in nature of the ...'s occuring lead to drastic divergence of the general notions. Most importantly, the solid condition in no way implies any kind of Hausdorff behavior. Essentially, the reason is that in the solid condition you already require in [...] a bunch of limits to exist uniquely (because you map in from a compact Hausdorff space). Then the solid condition is only that this generates more limits (and uniquely) by taking certain $\mathbb{Z}$-linear combinations. Whereas in the usual Cauchy completeness every limit that you posit exists, you also posit exists uniquely.
This phenomenon, that solid abelian groups incorporate non-Hausdorff behvaior, it absolutely crucial to having a functioning theory. Why? Because non-Hausdorff behavior is inevitable once you require an abelian category. If $M$ is some non-discrete condensed abelian group which you want to call "complete", and $N$ is any discrete dense subgroup mapping into $M$, the cokernel has to be "complete" as well, even though it's a classic example of a "bad quotient" which is non-Hausdorff.
(This is an example of the trade-off between "good categories of (mostly) bad objects" vs. "bad categories of good objects". But for me personally, I've seen enough examples of the solid formalism gracefully handling non-Hausdorff spaces in practice that I no longer think of them as "bad objects".)
You might be tempted to then think that solid abelian groups are more like weakened analog of complete topological abelian groups, where you drop the uniqueness requirement in the definition of completeness and only require existence of limits for Cauchy sequences. But no, the fact that solidness is an "exists a unique" condition is also crucial for it being an abelian category. "Exists" is just not stable enough a notion.
The other big difference is that a solid abelian group is "only required to be complete as far as compact subsets are concerned". If you have a Cauchy sequence which is not contained in a compact subset, the solid condition says nothing about it. Thus, for example, solid abelian groups are closed under arbitrary direct sums, whereas I don't think there's any reasonable topology on the direct sum $\oplus \mathbb{Z}_p$ for which it's complete [Edit: I thought wrong! See https://mathoverflow.net/questions/387322/countable-sum-bigoplus-n-0-infty-mathbb-z-p-as-a-topological-group]. Again, the fact that solid abelian groups are closed under all colimits is very important for us theoretically: it's part of what makes sure that the category of solid abelian groups "behaves like the category of modules over a ring" (formally, it is an abelian category generated by compact projective objects), and therefore has convenient algebraic properties.
Now, it seems the main thrust of your question as about defining possibile non-abelian analogs of solidness. I don't want to say this can't be done (I doubt it can but I certainly could be wrong), but I hope that the above remarks show that if you want to define such a notion, you shouldn't do it by trying to follow the usual presentation via Cauchy-completeness and uniform structures. Despite the fact that the two notions agree on many familiar objects, there is a huge divergence in general.
Best Answer
I would motivate them as follows: if topological spaces were invented to give a general meaning to "continuous function", then uniform spaces were invented to give a general meaning to "uniformly continuous function". It is clear what "uniformly continuous" should mean for metric spaces and topological groups, but how should the general notion be formalized?
Once this is formalized, one can define the notion of Cauchy net in a uniform space (which is something you cannot do for general topological spaces). This leads to the notion of completeness of course (every Cauchy net converges to at least one point), although the theory is much cleaner for complete Hausdorff uniform spaces, where you have convergence to at most one point as well.
To illustrate this: the Cauchy completion of a uniform space $X$ can be defined in the usual way via equivalence classes of Cauchy nets. It is a complete Hausdorff uniform space $\bar{X}$ together with a map $i: X \to \bar{X}$ which satisfies a universal property: given a complete Hausdorff uniform space $Y$ and a uniformly continuous function $f: X \to Y$, there is a unique uniformly continuous map $\bar{f}: \bar{X} \to Y$ such that $\bar{f} \circ i = f$. (If you omit "Hausdorff" or "uniformly", you lose the universal property, which is arguably the point of the completion.)
The nLab has an article on uniform spaces with some material not included in the Wikipedia article.