TLDR: Is there a way to talk about composition of paths in an arbitrary model category?
Intuition: In topological spaces, we can compose paths. That is, if $\gamma_1$ is a path from $x$ to $y$, and $\gamma_2$ is a path from $y$ to $z$, there is a composite path $\gamma_1 * \gamma_2$ that goes from $x$ to $z$. We can do this with our usual paths (maps from $[0,1]$ into $X$) or with Moore paths (maps from $[0,L]$ into $X$ for any $L≥0$).
Obviously there are multiple ways paths could be composed. For example, we could reparameterize each path and get a composite path with the same “image” but traversed in a different way. Path composition, as I see it, seems to have two import properties:
If $\gamma_1$ is a path from $x$ to $y$, and $\gamma_2$ is a path from $y$ to $z$, the composite path $\gamma_1 * \gamma_2$ is a path from $x$ to $z$;
If $C_x$ is a constant path (of length zero in the Moore case), we have $C_x * C_x = C_x$
I would like to show there is an analogous construction in a (very good) path space object in a model category.
My thoughts: Suppose $\mathcal{M}$ is a category with a model structure $(\mathcal{C}, \mathcal{W}, \mathcal{F})$. Let $X$ be an object of $\mathcal{M}$. Take a very good path space object for $X$. That is, factor the diagonal $\Delta_X : X \rightarrow X \times X$ $$X \xrightarrow{r} P_X \xrightarrow{\epsilon} X \times X$$ where $r$ is an acyclic cofibration, and $\epsilon$ is a fibration. We may make a space of composable paths $CP_X$ by the following pullback square. $$\require{AMScd} \begin{CD} CP_X @>{p_1}>> P_X\\ @V{p_0}VV @VV{\pi_0\epsilon}V \\ P_X @>{\pi_1\epsilon}>> X \end{CD}$$ We think of this pullback as pairs of paths (picked out by $p_0$ and $p_1$) such that the right endpoint of the first path is the left endpoint of the second path. A composition operation should be a map $\mathbf{c}:CP_X \rightarrow C_X$. One way to find such a $\mathbf{c}$ with the desired properties would be to solve the following lifting problem. $$\require{AMScd} \begin{CD} X @>{r}>> P_X\\ @V{R }VV @VV{\epsilon}V \\ CP_X @>{\langle \pi_0 \epsilon p_0, \pi_1 \epsilon p_1 \rangle}>> X\times X \end{CD}$$ $R:X \rightarrow CP_X$ is the map induced by the universal property of pullbacks and the following square. $$\require{AMScd} \begin{CD} X @>{r}>> P_X\\ @V{r}VV @VV{\pi_0\epsilon}V \\ P_X @>{\pi_1\epsilon}>> X \end{CD}$$ A solution to this lifting problem would be an arrow $\mathbf{c}:CP_X \rightarrow P_X$ such that $\mathbf{c} \circ R = r$ and $\epsilon \circ \mathbf{c} = \langle \pi_0 \epsilon p_0, \pi_1 \epsilon p_1 \rangle$. This first equality would say that composing two of the paths from $r$ just gives you the path from $r$ back. The second equality says that we have the correct endpoints. This would be exactly what we want.
If the map $R$ is an acyclic cofibration, we get a lift from the model structure. I don’t see how to prove that $R$ is an acyclic cofibration though. I would happily accept an answer that adds extra hypotheses.
My main answer.
I will first propose an answer to your question. After that, I write down some afterthoughts that may be interesting, even though they do not help answering your question.
So, as for your actual question: you will need some extra conditions on your model category in order to guarantee that $R$ is a cofibration.
I propose only working with fibrant $X$, because the pullback diagram defining $P_X$ will otherwise generally not be a homotopy pullback, and I think we would want to make path composition homotopy invariant.
Assumption 1: The object $X$ is fibrant in $\mathcal{M}$.
Now the map $R$ is always a weak equivalence. This is because $(\pi_0\varepsilon)r=\mathrm{id}_X$, so $\pi_0\varepsilon$ is a weak equivalence. Moreover, it is a fibration as composition of fibrations (we use that $X$ is fibrant in order to conclude that $\pi_0\colon X\times X\to X$ is a fibration). Therefore $p_0\colon CP_X\to P_X$ is an acyclic fibration as pullback of the acyclic fibration $\pi_0\varepsilon$. The relation $r=p_0R$ now gives you that $R$ is a weak equivalence.
To show that $R$ is a cofibration, I will basically cheat, and assume the following.
Assumption 2: Suppose given a commutative diagram $$ \require{AMScd} \begin{CD} A @>>> B @<<< C\\ @VVV @VVV @VVV\\ A' @>>> B' @<<< C' \end{CD} $$ in $\mathcal{M}$, in which the horizontal arrows are fibrations and the vertical arrows are cofibrations. Then the induced map $A\times_BC\to A'\times_{B'}C'$ is a cofibration.
The map $R$ equals $X\cong X\times_XX\xrightarrow{r\times r}PX\times_XPX\cong CP_X$, so using that $p_0$ and $p_1$ are fibrations (since $X$ is fibrant), this second assumption immediately implies that $R$ is a cofibration.
In a way, I really don't think you can avoid a strong assumption like this, because $R$ really is just a map induced on pullbacks by cofibrations, and in general model categories this will not be a cofibration. Luckily, there are quite a few situations in which Assumption 2 holds. For instance, it holds for the Quillen model structure on convenient topological spaces, but also for the Strøm model structure on the category of all topological spaces. (For the latter, see for instance R.W. Kieboom's paper A pullback theorem for cofibrations, https://link.springer.com/content/pdf/10.1007/BF01165895.pdf. )
Another class of examples in which Assumption 2 holds is given by those model categories in which the cofibrations are precisely the monomorphisms (which holds for instance for the Kan model structure and Joyal model structure on $\mathrm{sSet}$, the Reedy/injective model structure and the (complete) Segal space model structure on $\mathrm{sSet}^{\Delta^\mathrm{op}}$, and the injective model structure on nonnegative cochain complexes or the projective model structure on chain complexes of $k$-vector spaces for a field $k$). It is in this case even easier to show that $R$ is a cofibration, because the composition $X\xrightarrow{R} CP_X\xrightarrow{p_0} P_X$ equals the cofibration $r\colon X\to P_X$ and is hence a monomorphism. Therefore $R$ is a monomorphism, so that it is a cofibration.
I don't expect $R$ to be an acyclic cofibration for general fibrant $X$ in a general (nontrivial) family of model categories that doesn't satisfy Assumption 2, but if anyone has suggestions, I am definitely interested.
Side remark.
In the $\infty$-category $\mathcal{M}_\infty$ underlying the model category $\mathcal{M}$, the lifting problem can always be solved. This is actually trivial: since $R\colon X\to CP_X$ is an equivalence in $\mathcal{M}_\infty$, you can informally just take the lift to be $rR^{-1}\colon CP_X\to P_X$ (and can obtain the homotopies witnessing commutativity without much effort). In a sense, this means that your question is not intrinsically homotopically interesting (of course, we sort of already knew that, because the unit interval in topological spaces (or simplicial sets) is contractible): since $r\colon X\to P_X$ is an equivalence, our ''composition map'' $CP_X\to P_X$ is equivalent to the identity on $X$ and therefore all $\infty$-categories have this path composition map built in in their identity maps. However, you are asking whether we can ''strictify'' this path composition construction, in the sense that you want to witness this in a homotopically meaningful way in the model categories. Usually I just try to work in $\infty$-categories, but that is not the right approach here because your question is specifically about the framework of model categories.