In a ring, result of multiple (of "addition" operation) is not the same as result of multiplication, correct? In a ring, a multiple for addition is written as $na$ to stand for $(a + a + ... + a)$.
This is not necessarily the same as $n * a$ (the "multiplication" operation).  Is that correct?
Multiple is only the same as multiplication for specific rings such as Integers.  Is that right?
I suspect the answer to be the case but I have never seen a proof one way or the other.
Thanks
 A: Addition and multiplication distribute in a ring.
If we start with
$$x+x+x+x$$
in a ring $R$, and we assume $1_R$ is an element of your ring (the multiplicative identity, which most definitions of Ring assume is there), we get:
$$1_R*x + 1_R*x + 1_R*x + 1_R*x$$
$$(1_R+1_R)*x + (1_R+1_R)*x$$
$$(1_R+1_R+1_R+1_R)*x$$
$$4_R*x$$
where we define $4_R$ as an element of the ring to be $1_R+1_R+1_R+1_R$.
Note that this works equally well with right-multiplication, so $4_R*x = x+x+x+x = x*4_R$.
For elements of your ring that cannot be found through repeated addition of $1_R$, left-multiplication and right-multiplication by them may not be the same.  $2\times2$-matrices furnish an example.
Even if multiplication is commutative, the above argument may fail. $R=\mathbb{Z}_3^2$ (2-tuples of integers mod 3) with element-wise $+$ and $*$ has $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}$ as elements. Here $1_R$ is $(1,1)$, $0_R$ is $(0,0)$, $3_R$=$0_R$ and $(2,1)*(1,2)=(2,2)$ while no sum of either $(2,1)$ or $(1,2)$ adds to $(2,2)$.
A: Let me provide an alternate answer, since I understood your question a little differently than the other answers. Here is what you asked:

In a ring, a multiple for addition is written as $na$ to stand for $(a + a + ... + a)$.
This is not necessarily the same as $n * a$ (the "multiplication" operation).  Is that correct?

That's not correct -- they are both the same thing! To clarify, the second notation only makes sense in a ring with unity / multiplicative identity, but as long as we are in such a ring, we must be clear: the definitions of these two notations are equivalent.
In more detail: for $n$ an integer and $r$ a ring element in an arbitrary ring $R$, we define
$$
nr =_{\text{def}} \underbrace{r + r + \cdots + r}_{n \text{ times}}.
$$
Now, if $r$ has a multiplicative identity $e$, then we also define, for $n$ an integer,
$$
n =_{\text{def}} \underbrace{e + e + \cdots + e}_{n \text{ times}}.
$$
Now, by distributivity,
$(e + e + \cdots + e)* r = r + r + \cdots + r$.
In other words, if we do $n * r$ with ring multiplication or with repeated addition, we get the same answer.
This is good -- we don't want our two notations to conflict.
As other answers have mentioned, in a general ring we might have $n = 0$ for some positive integer $n$. But that's really irrelevant, because the two notations still coincide even in this case.

P.S. The most important thing is to always keep separate in your head which variables are integers (like $n$, $0$, $1$) and which variables are ring elements (like $r \in R$). It gets confusing since the integers themselves are a ring, and since we have these notations such as $n * r$ for repeated addition, and since $1$ can be used both for a ring element and for an integer, etc. But as long as you are keeping the set of integers separate in your mind from the ring, you should be able to avoid confusion.
A: For any ring $R$ (not even with identity, necessarily) and $r\in R$ and any positive integer $n$, $n\cdot r$ is defined to be $r+r+\ldots$ $n$ times. This extends to negative numbers and $0$ in the obvious ways. 
So yes, it is that way for the product of an element with an integer.
The place where the phrase stops making sense is the product of any two elements of a ring. For instance, in $F[x]$, how do you add $x$ to itself $x$ times?
Under the hood, the notation you're talking about is just the notation for repeated group operations. If you write a group multiplicatively, then taking $n$ copies of $g$ via the operation is written as $g^n$ when the operation is written multiplicatively, and is written $ng$ when the group is written additively.
A: Note that  in your  question the symbol $n$ is used with two different meanings.

In  the statement:
  
  
*
  
*In a ring, a multiple for addition is written as $\color{blue}{na}$ to stand for $(a + a + ... + a)$.
  
  
  the symbol $n$ denotes the natural number $n\in\mathbb{N}$ which is used to indicate a multiple of the ring-element $a$. 
  \begin{align*}
\underbrace{a+a+\cdots+a}_{\color{blue}{n}\text{ times}}\qquad\qquad  n\in\mathbb{N}
\end{align*}

It is clever to write it stands for instead of $na=a+a+\cdots+a$, since these elements  are not necessarily comparable (the ring $R$ does not necessarily contain natural numbers).

On the other hand, in the statement
  
  
*
  
*This is not necessarily the same as $\color{blue}{n \ast a}$ (the "multiplication" operation). 
  
  
  the symbol $n$ denotes a completely different object, namely an element of the ring $R$.

Keeping this in mind the answer to your question is yes, your assumption is correct. We just need to consider a ring which don't contain natural numbers.

Example: We consider the ring $R=\left(M_2(\mathbb{R}),+,\ast\right)$ of all real $(2\times 2)$-matrices with the usual matrix addition and multiplication.
For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in R$ and $n\in\mathbb{N}$ we obtain
  \begin{align*}
nA=n\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}na&nb\\nc&nd\end{pmatrix}\in R
\end{align*}
  but $n\ast A$ is not defined since $n\not\in R$.

A: Let's use two different notations: $n\cdot a$ for repeated addition and $a*b$ for ring multiplication, so that we can discuss when it makes sense to compare them. And when it does make sense to do so, when are they equal.
In a ring $n\cdot a$, for $n$ a natural number, refers to repeated addition: $$n\cdot a=\underbrace{a+\dotsb+a}_{n\text{ times}}.$$ Using additive inverse, we may also define the expression for $n$ any integer: $(-n)\cdot a=-(n\cdot a)$.
The ring also has a multiplication operation $a* b$ which has nothing to do with repeated addition; you can't add $b$ to itself $a$ times if $a$ is not a counting number. Rather multiplication is an operation given as part of the given structure of the ring, separate from addition.
It does not make sense to ask whether $a * b$ and $a\cdot b$ are the same, because the latter expression is undefined for a general ring element. The question is only even defined for expressions of the form $n\cdot b$, where $n$ is an integer.
If the ring has a multiplicative identity element $1$, which most authors do assume, then the numbers $$n=\underbrace{1+1+\cdots+1}_{n\text{ times}}$$ may also be viewed as ring elements. (But beware that in some rings we may have $\underbrace{1+1+\cdots+1}_{n\text{ times}}=0$ for $n\neq 0.$) These elements make a subring of $R$ called the prime ring (see the second definition).
So in rings with identity, multiplication of general ring elements may be viewed as an extension of multiplication of elements of the prime subring. Meaning that when if a ring element $a$ is equal to $n$ for some integer $n$, then $a*b=a\cdot b$. So the answer is that your proposed phrasing "$n*a$ is not necessarily the same as $na$" is not correct; $n*a$ and $n\cdot a$  are necessarily the same. As functions, the operation $\cdot$ is a subset of $*$. Note that multiplication is always commutative on the prime subring $m\cdot n=n\cdot m$, whereas it need not be on the whole ring $a* b\neq b* a.$
In a ring without identity element, (sometimes called rngs. Get it? Remove the 'i' from "rings"), both types of multiplication exist together, but multiplication by integers is not a subset of the ring's multiplication operation. Note that a ring without identity can be canonically embedded in one which does include identity.
So to sum up, comparing $n * b$ and $n\cdot b$ only makes sense when $n$ is an integer, not a general ring element and only if our ring contains a multiplicative identity. In that case, they must agree.
A: Any ring $R$ is set equipped with two binary operations addition and multiplication. Here we required addition to be commutative (additive abelian group). Thus any ring is an $\Bbb{Z}$-module and therefore your intuition is correct. 
A: None of the other answers clearly stated what I was looking for when I found this question. Namely, if $(R, +, *)$ is a ring (with multiplicative identity $1_R$), then for all $n \in \mathbb{Z}_{\geq 1}$ and $r \in R$,
$$\begin{align*} 
n \cdot r 
&= \underbrace{r + \cdots + r}_{n\text{ times}} \tag{definition of $n \cdot r$} \\
&= \underbrace{1_R * r + \cdots + 1_R * r}_{n\text{ times}} \\
&= \left( \underbrace{1_R + \cdots + 1_R}_{n\text{ times}} \right) * r \tag{right distributivity} \\
&= \left( n \cdot 1_R \right) * r \\
&= n_R * r \tag{define $n_R := n \cdot 1_R$}
\end{align*}$$
If we instead write $\underbrace{r + \cdots + r}_{n\text{ times}} = \underbrace{r * 1_R + \cdots + r * 1_R}_{n\text{ times}}$ and use left distributivity, we get $n \cdot r = r * n_R$.
Additionally,
$$\begin{align*} 
(-n) \cdot r 
&= -(n \cdot r) \\
&= -(n_R * r) \\
&= -1_R * (n_R * r) \\
&= (-1_R * n_R) * r \\
&= (-(n_R)) * r \\
&= (-n)_R * r 
\end{align*}$$
since $(-n)_R := (-n) \cdot 1_R = - (n \cdot 1_R) = - (n_R)$. Similarly,
$$(-n) \cdot r = n \cdot (-r) = (-r) * n_R = (r * -1_R) * n_R = r * (-(n_R)) = r * (-n)_R$$
With $0$ denoting the additive identity, we define $0_R := 0 \cdot 1_R = 0$. Thus $0 \cdot r = 0 = 0 * r = 0_R * r = r * 0_R$. This brings us to the pleasing conclusion that for all $m \in \mathbb{Z}$ and $r \in R$,
$$m \cdot r = m_R * r = r * m_R$$
