Homology and cohomology of free loop spaces String topology, as well as Hochschild (co)homology, give a rich perspective on the homology and cohomology of a free loop space $LM$ of a manifold $M$.
Let $k$ be a field and let $M$ be $n$-dimensional.
There is a graded ring $H^*(LM, k)$ with multiplication the usual multiplication on cohomology of a space. If $M$ is oriented, there is a graded ring $\mathbb{H}_*(LM) = H_{*+n}(LM, k)$ with multiplication given by the Chas-Sullivan product.
Is there any relation between these graded rings? One should feel free to assume that $M$ is simply connected and also that $TM$ is stably trivial if that improves the situation. Of course, in that case one has that $H_*(LM) = HH_*(C^*(M), C^*(M))$ while $\mathbb{H}_*(LM) = HH^*(C^*(M), C^*(M))$  (where $C^*(M)$ denotes the algebra of cochains on $M$) but from the fact perspective the cup-product structure on $H^*(LM)$ is wholly not self-evident (for $k = \mathbb{Q}$, where $C^*(M)$ can be made into a cdga, it arises from the product on $C^*(M)$ and the shuffle product).
 A: I'll address your question for a reason why the Hochschild homology $HH_*(C^*(M),C^*(M))$ should have a cup product, without resorting to the fact that it is the cohomology of a space. We assume everything implicitly rationalized and $M$ simply connected.
The Hochschild homology of $C^*(M)$ can be computed as the factorization homology
$\int_{S^1} C^*(M)$. The relation between Hochschild homology and the free loop space is an artifact of the more homotopical claim that $\int_{S^1} \Sigma^\infty M^\vee \simeq \Sigma^\infty_+ \operatorname{Map}(S^1,M)$. So to address your question, we should supply $\int_{S^1} \Sigma^\infty M^\vee$ with a diagonal without actually computing the factorization homology.
Factorization homology over $N$ can be computed as a bar construction of the little disks modules $E_N$ with an $E_n$ algebra $A$. Let $(S^1)^\times$ denote the right $\mathrm{com}$ module which in degree $i$ is $(S^1)^{\times i}$ with right module structure induced by the diagonal. This is equivalent to $\mathrm{ind}^{\mathrm{com}}_{E_1} E_{S^1}$. We need 3 facts: the zLie algebra model of $M$ is $K(\Sigma^\infty M^\vee)$, the enveloping algebra of this is $\Sigma^\infty \Omega M$, and $K(\Sigma^\infty_+(S^1)^\times) \simeq \Sigma^\infty_+ E_{S^1}$ as $\mathrm{lie}$ modules. The first two are classical, and the third is a consequence of the Koszul self duality of the maps of operads $\mathrm{lie} \rightarrow \mathrm{Ass} \rightarrow \mathrm{com}$. Combined with basic facts about induction and restriction, we may compute:
\begin{align}\int_{S^1} \Sigma^\infty_+ M ^\vee &\simeq B\big(\Sigma^\infty_+E_{S^1},\Sigma^\infty_+E_1, \Sigma^\infty M^\vee\big)\\
&\simeq 
B\big(\mathrm{ind}_{E_1}^{\mathrm{com}} \Sigma^\infty_+E_{S^1},\mathrm{com}, \Sigma^\infty M^\vee\big)\\
&\simeq B\big(\Sigma^\infty_+(S^1)^\times,\mathrm{com},\Sigma^\infty M^\vee \big)\\
&\simeq 
B\big(K(\Sigma^\infty_+(S^1)^\times),K(\mathrm{com}),K(\Sigma^\infty M^\vee\big)\\
&\simeq B\big(\Sigma^\infty_+ E_{S_1},\Sigma^\infty_+E_1,\mathrm{ind}_{\mathrm{lie}}^{E_1}  K(\Sigma^\infty M^\vee)\big)\\
&\simeq  B\big(\Sigma^\infty_+ E_{S_1},\Sigma^\infty_+ E_1,\Sigma^\infty \Omega M\big)\simeq \int_{S^1} \Sigma^\infty \Omega M
\end{align}
This last algebra has a diagonal since it is the suspension spectrum of a space, hence the factorization homology has one as well. Note: I have played fast and loose with suspensions, so some of these equivalences might only hold plus or minus some desuspensions.
