Free algebra over an operad is an algebra over that operad Let us focus on operads of vector spaces and let $V$ be a vector space and $P$ an operad. The free $P$-algebra on $V$ is defined by $P(V)=\bigoplus_{r=0}^\infty (P(r)\otimes V^{\otimes r})_{\Sigma_r}$, where the $Σ_r$-quotient identifies tensor permutations with the action of permutations on $P(r)$.
On the other hand, $V$ is said to be a $P$-algebra if there is a morphism of operads $P\to End_V$, where $End_V$ is the endomorphism operad of $V$. Equivalently, $V$ is a $P$-algebra if there is a collection of maps $P(r)\otimes V^{\otimes r}\to V$ satisfying certain conditions.

How do these two notions reconcile?

An element of $p\otimes x_1\otimes\cdots \otimes x_r\in P(V)$ can be written as $p(x_1\otimes\cdots\otimes x_r)$ and therefore $p$ is interpreted as a map $V^{\otimes r}\to V$. But how can we realize $p$ as an element of $End_V(r)$ so that we do have the map of operads $P\to End_V$? Or equivalently, how can identify $p(x_1\otimes\cdots\otimes x_r)$ with an element of $V$ so that we have the maps $P(r)\otimes V^{\otimes r}\to V$?
I am also interested in the case that $V$ is a dg-module, so I'm hoping for an answer that extends to dg-modules as well.
 A: Let me deal with the case of a nonsymmetric operad for simplicity (so we forget about $\Sigma_r$). If you have symmetric operads, it's essentially the same story, but there are more things to write down
(note : I'm assuming we're dealing with a unital operad, otherwise the terminology "free algebra on $V$ is not super well-suited)
Let $(C,\otimes, 1)$ be a symmetric monoidal cocomplete category such that the tensor product commutes with colimits in each variable.
You may think of $C=\mathsf{Vect},  \mathsf{Ch}$ (chain complexes over a commutative ring, e.g. $\mathbb Z$ or a field $k$), or $\mathsf{sSet}$ for instance, or even $\mathsf{Set}$ for more basic examples; and let $P$ be an operad in $C$.
Let $V\in C$, and let $X$ be a $P$-algebra in $C$, and suppose $V\to X$ is a morphism in $C$. Then, for each $r$, you get a map $P(r)\otimes V^{\otimes r} \to P(r)\otimes X^{\otimes r}$ which is naturally defined, and, since $X$ is an algebra, you can compose it with its structure maps to get a map $\mu_r : P(r)\otimes V^{\otimes r}\to X$.
Note that for each $n_1,...,n_r$ adding up to $n$, you get that the two maps
$P(r)\otimes P(n_1)\otimes... \otimes P(n_r)\otimes V^{\otimes n}\rightrightarrows X$
(defined respectively by $P(r)\otimes P(n_1)\otimes ... \otimes P(n_r)\to P(n)$ followed by $\mu_n : P(n)\otimes V^{\otimes n}\to X$; and $P(n_1)\otimes V^{\otimes n_1}\otimes ... \otimes P(n_r)\otimes V^{n_r}\overset{\mu_{n_1}\otimes ...\otimes \mu_{n_r}}\to X^{\otimes r}$ followed by $P(r)\otimes X^{\otimes r}\to X$)
agree. I'll let you understand why that is (it relies on the axioms for algebras over $P$ that $X$ satisfies)
All in all, we get a map $\bigoplus_{r\geq 0}P(r)\otimes V^{\otimes r}\to X$ in $C$; this map is moreover a map of $P$-algebras if we let $\bigoplus_{r\geq 0}P(r)\otimes V^{\otimes r}$ have the "tautological structure", defined by :
$$P(n)\otimes (\bigoplus_{r\geq 0}P(r)\otimes V^{\otimes r})^{\otimes n} \cong P(n)\otimes \bigoplus_{r_1,...,r_n}P(r_1)\otimes ... \otimes P(r_n)\otimes V^{\otimes\sum r_i}\\
\cong P(n)\otimes \bigoplus_{k\geq 0}\bigoplus_{r_1,...,r_n, \sum r_i = k}P(r_1)\otimes ... \otimes P(r_n)\otimes V^{\otimes k}\\
\cong \bigoplus_{k\geq 0}\bigoplus_{r_1,...,r_n, \sum r_i = k} P(n)\otimes P(r_1)\otimes ... \otimes P(r_n)\otimes V^{\otimes k}\\
\to \bigoplus_{k\geq 0}P(k)\otimes V^{\otimes k}$$
where all but the last line are just rearrangements of terms using the fact that $\otimes$ commutes with direct sums in each variable, and the last line is given by the structure maps of $P$.
Those give us maps $P(n)\otimes P(V)^{\otimes n}\to P(V)$, and one checks that this gives $P(V)$ a $P$-algebra structure; for which the above map $P(V)\to X$ is a $P$-algebra map.
This construction establishes a natural isomorphism $\hom_C(V,UX)\cong \hom_{\mathsf{Alg}_P}(P(V),X)$, where $UX$ denotes the underlying $C$-object of $X$.
(there are things to check here, for instance that's where I'd use unitality of $P$, to construct the inverse map)
This is what is meant by "$P(V)$ is the free $P$-algebra on $V$" (the technical term is that $V\mapsto P(V)$ is left adjoint to the forgetful functor $\mathsf{Alg}_P\to C$)
A more down-to-earth description of this structure on $P(V)$ (in the case where we have elements, e.g. in $\mathsf{Vect},\mathsf{Set}$) is given by $p\otimes (q_1\otimes x^1_1 \otimes ... \otimes x^1_{r_1} \otimes ... \otimes q_n\otimes x^n_1\otimes...\otimes x^n_{r_n}) \mapsto \mu(p,q_1,...,q_n)\otimes x^1_{r_1}\otimes...\otimes x_{r_n}^n$ where $\mu : P(n)\otimes P(r_1)\otimes ... \otimes P(r_n)\mapsto P(\sum r_i)$ is the structure map of $P$.
One way to think about this is similar to free groups (where each element of the free group is a word on the elements of the set you started with, maybe with inverses (those correspond to the operations in $P$), and where multiplication is just given by concatenation (here you simply have operations to add to the mix, instead of just concatenation)
Now if you're considering a symmetric operad, everything is the same, except that in order for $P(V)$ to satisfy the $\Sigma_r$-equivariance axioms, you have to mod out by the $\Sigma_r$-action. The details can be tedious to write down, so I'll leave that as an exercise to you (be warned that once you understand the nonsymmetric case, dealing with those $\Sigma$-details is not super-enlightening)
So, for a tldr; :
$P(V)$ comes with an algebra structure, which is essentially "tautological" : it is induced by the structure maps of $P$ itself; it's not the same algebra structure as one on $V$ if it already has one. In particular it is "free" on $V$ : a map of algebras $P(V)\to X$ is the same data as a map in $C$: $V\to X$.
An algebra structure on $V$ can be specified by a map (in $C$) $P(V)\to V$ satisfying some conditions (look up "monad" if you want to see what those are)
