Clarifications needed in the definition in making a module into an algebra I found the following definition for how to make a module over a ring into an algebra:

Let $R$ be a commutative unitary ring.  By an $R-$algebra we shall mean an $R-$module $A$ together with an internal law of composition $A\times A\mapsto A$, described by $(x,y)\mapsto xy$ and called $\textit{multiplication,}$ which is distributive over addition and such that 
$(\forall\lambda\in R)(\forall x,y\in A) $ $\lambda(xy)=(\lambda x)y=x(\lambda y)$

The author then further states:  

"By imposing conditions on the multiplication in the above definition we obtain various types of algebra.  For example, if the multiplication is associative then $A$ is called an $\textit{associative algebra}$ (note that in this case $A$ isa ring under its internal laws of addition and multiplication); if the multiplication is commutative then $A$ is called a $\textit{commutative algebra};$ if there is a multiplicative identity element in $A$ then $A$ is said to be $\textit{unitary}.$  A unitary associative algebra in which every non-zero element has an inverse is called a $\textit{division algebra}.$"

I have a few questions about the common definition of an algebra use in the context of modules.
(1) when a module say $M$ defined over a ring $R$, the module $M$ is called a vector space over the ring $R$ even though $R$ might not be a field?
(2) The definition of an algebra for a module makes the ring $R$ acting on $M$ a bilinear map?
(3) when speaking of a module $M$ being either an associative or a commutative or a unital algebra and needing to satisfied: $(\forall\lambda\in R)(\forall x,y\in A) $ $\lambda(xy)=(\lambda x)y=x(\lambda y)$, does it mean for associativity:
$\forall\lambda\in R$ and $\forall x, y, z \in M$ $\lambda x(yz)=\lambda (xy)z$ or does it mean  $\forall\lambda\in R$ and $\forall x, y \in M$ $\lambda (xy)=(\lambda x)y$
For being commutative, which one of the following does it mean:
$\forall\lambda\in R$ and $\forall x, y \in M$ $\lambda (xy)=\lambda (yx)$ or does it mean  $\forall\lambda\in R$ and $\forall x\in M$ $\lambda x=x\lambda$
Lastly the module $M$ being an unital $R-$algebra, since $M$ is already an additive abelian group, does both $M$ and $R$ has to have its own identity elements (multiplicative for $M$ or additive or multiplicative for $R$) or is it only that one of $M$ or $R$ needs to have both an additive and a multiplicative identity.   Thank you in advance.
 A: 
when a module say $M$ defined over a ring $R$, the module $M$ is called a vector space over the ring $R$ even though $R$ might not be a field?

No, $M$ would only be an $R$-vector space if $R$ is a field. That said, it's definitely possible for someone to accidentally refer to a module as a vector space. Many results about vector spaces generalise to modules - for example, a linear map $R^n \to R^n$ is invertible if and only if the determinant of the relevant matrix is a unit.
Many applications of algebras are when the ring is a field. For example, the Stone-Weierstrass theorem deals with $\mathbb{R}$-algebras (which are, of course, vector spaces).

The definition of an algebra for a module makes the ring $R$ acting on $M$ a bilinear map?

In fact, we only need $M$ to be a module for the multiplication function $(r \in R, m \in M) \mapsto r \cdot m$ to be bilinear. This is one of the axioms of a module. So an algebra structure on $M$ is not relevant here.
Associativity here means that $\forall a, b, c \in M (a(bc) = (ab)c)$. Your first statement $\forall \lambda \in R \forall x, y, z \in M (\lambda (x(yz)) = \lambda ((xy) z)$ is equivalent to this claim (in particular, in the case where $\lambda = 1$). Your second statement $\forall \lambda \in R \forall x, y \in M (\lambda (xy) = (\lambda x) y)$ is part of the definition of an algebra.
For being commutative, what is meant is that $\forall a, b \in M (ab = ba)$. Your first statement $\forall \lambda \in R (\forall x, y \in M (\lambda (xy) = \lambda (yx))$, is equivalent to commutativity. Your second statement, $\forall \lambda \in R \forall x \in M (\lambda x = x \lambda)$ doesn't make any sense, since $x \lambda$ does not make any sense.
Being unitary simply means there is some $e \in M$ such that $\forall x \in M (x = ex = xe)$. Note that a unit, if it exists, is unique, since if we had units $e_1$ and $e_2$ then $e_1 = e_1 e_2 = e_2$. Generally, we would only require the algebra to be unitary if $R$ is also unitary.
If $R$ is a (not necessarily commutative) unital ring, then we can describe an associative unital $R$-algebra as some (not necessarily commutative) unital ring $M$ together with a ring homomorphism $i : R \to M$. It's easy to show that every unital associative $R$-algebra $M$ is a ring with the ring homomorphism map $i(x) = x \cdot 1$ (where $1$ is the unit in $M$); conversely, the other direction gives us an $R$-algebra from a ring $M$ together with $i : R \to M$ through the module structure $(r, m) \mapsto i(r) \cdot m$. Of course, the requirement that $M$ moreover be a commutative algebra is easily expressed as requiring $M$ as a ring to be commutative.
