I think you will see the difference if we explicitly write out the definitions for continuity and uniform continuity.
Let $f: X \rightarrow \mathbb{R}$ be a function from some subset $X$ of $\mathbb{R}$ to $\mathbb{R}$.
Def 1: We say that $f$ is continuous at $x$ if for every $\epsilon > 0$ there exists $\delta > 0$ such for every $y \in X$, if $|y - x| < \delta$ then $|f(y) - f(x)| < \epsilon$.
Def 2: We say that $f$ is continuous if $f$ is continuous at every $x \in X$.
This definition corresponds to what you called "continuity on a set" in your question.
Def 3: We say that $f$ is uniformly continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that for any two points $x, y \in X$, if $|y - x| < \delta$ then $|f(y) - f(x)| < \epsilon$.
As you have already noted in your question, the difference between uniform continuity and ordinary continuity is that in uniform continuity $\delta$ must depend only on $\epsilon$, whereas in ordinary continuity, $\delta$ can depend on both $x$ and $\epsilon$.
To illustrate this with a concrete example, let's say $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$, and $f(10) = 15, f(20) = 95$ and $f(30) = 10$.
In uniform continuity, if I pick $\epsilon > 0$, you must give me a single $\delta > 0$ such that the following statements all hold:
- If $|y - 10| < \delta$ then $|f(y) - 15| < \epsilon$.
- If $|y - 20| < \delta$ then $|f(y) - 95| < \epsilon$.
- If $|y - 30| < \delta$ then $|f(y) - 10| < \epsilon$.
On the other hand, in ordinary continuity, given the same $\epsilon > 0$ as above, you are free to pick three different deltas, call them $\delta_1, \delta_2, \delta_3$ such that the following statements all hold:
- If $|y - 10| < \delta_1$ then $|f(y) - 15| < \epsilon$.
- If $|y - 20| < \delta_2$ then $|f(y) - 95| < \epsilon$.
- If $|y - 30| < \delta_3$ then $|f(y) - 10| < \epsilon$.
To answer the second part of your question, as Jacob has already pointed out, your statement is the sequential characterisation of ordinary continuity. Since ordinary continuity does not imply uniform continuity the "if" part of your statement is false. However, it is easy to see that uniform continuity implies ordinary continuity, so the "only if" part of your statement is true.
First of all, continuity is defined at a point $c$, whereas uniform continuity is defined on a set $A$. That makes a big difference.
But your interpretation is rather correct: the point $c$ is part of the data, and is kept fixed as, for instance, $f$ itself. Roughly speaking, uniform continuity requires the existence of a single $\delta>0$ that works for the whole set $A$, and not near the single point $c$.
Best Answer
Here's a picture that might help. A visual way of understanding $\delta$-$\epsilon$ arguments is by starting with a $\delta$-sized area in the domain, projecting up to the function, and then back onto an $\epsilon$-sized area in the range.
With the function $f(x) = x$, there is a bounded ratio between the size of the area I feed in and the size of the area I get out. Not so with $f(x) = x^2$! Look how I feed in small areas and get out large areas for large values of $x$. This is why we say that $f(x) = x$ is uniformly continuous, but $f(x) = x^2$ is not uniformly continuous on $\mathbb{R}$. There's no way to globally (i.e. independent of $x$) control the size of the image of $f(x) = x^2$ by controlling the size of the domain.