The notion of free algebra functor I am new in the category theory. According to the definition of functor, a functor is a type of mapping between categories arising in category theory. Functors can be thought of as homomorphisms between categories. During studying the notion of Algebraic Operad we have the following definition:
Given a type of algebras the algebraic operad is given by the functor
free algebra, which is a monad in Vect. If the relations are multilinear, then the endofunctor is a Schur functor.
What is the meaning of the functor free algebra in the above definition?
 A: Given a category of algebraic objects $C$, such as groups, rings, modules, etc., equipped with the forgetful functor $U : C \to \text{Set}$ sending each object to its underlying set, the free algebra functor is the left adjoint of $U$; this means that there is a natural isomorphism
$$\text{Hom}_C(F(S), X) \cong \text{Hom}_{\text{Set}}(S, U(X))$$
for all $S \in \text{Set}, X \in C$. For groups this is the free group functor, etc. 
$U$ and $F$ together give rise to a functor $M = U \circ F : \text{Set} \to \text{Set}$ which has the structure of a monad. For algebraic objects the adjunction above is typically monadic, meaning $C$ can be recovered from $M$ in a particularly nice way; in this case $M$ encodes the algebraic theory describing the objects of $C$. 
If $C$ consists of algebras of some sort over vector spaces (e.g. algebras, commutative algebras, Lie algebras) then we can instead consider a forgetful functor $U : C \to \text{Vect}$ to vector spaces and a corresponding left adjoint $F : \text{Vect} \to C$. If $C$ is particularly nice it can be described by an operad $O$, in which case the induced monad $M = U \circ F : \text{Vect} \to \text{Vect}$ takes the form
$$M(V) = \bigoplus_{n \ge 0} O_n \otimes_{S_n} V^{\otimes n}$$
for some sequence of vector spaces with $S_n$-action $O_n$ making up the operad $O$. Functors of this form are sometimes called Schur functors, although that term is also sometimes reserved for special cases. 
There's a lot of material implicit in these two sentences, and especially if you're new to category theory it could be awhile before you can fully understand them. 
