Ideal Generated by a Finite Number of Polynomials? Background
I have been confused about a particular definition in the textbook for my abstract algebra class, Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea.  It is frustrating because I feel that I have a partial grasp on what the definition is trying to say, but when it comes down to it I simply find my confused.  So, instead of suffering by myself for any longer with this definition, I have decided to turn to you lovely folks to help me understand this seemingly simple concept.

What I Understand
First off, I understand (at least in the context of this book) what an ideal is.  The definition my book gives for an ideal is

Definition.  A subset $I\subseteq k[x_1,..., x_n]$ is an ideal if it satisfies:
(i)  $0\in I$.
(ii) If $f,g\in I$, then $f+g\in I$.
(iii)  If $f\in I$ and $h\in k[x_1,...,x_n]$, then $hf\in I$.

I find this to be a simple, easy to understand definition.  My problem, however, arises a few lines later when they define an ideal generated by a finite number of polynomials.

What I Don't Understand
And now, I give you the definition that has been causing me an incredible amount of confusion and frustration.

Definition.  Let $f_1,...,f_s$ be polynomials in $k[x_1,...,x_n]$.  Then we set $$\langle f_1,...,f_s \rangle=\Big\lbrace \sum_{i=1}^s h_if_i \ | \ h_1,...,h_s\in k[x_1,...,x_n] \Big\rbrace.$$

I know what you are thinking.  How does he not understand this?  I wish I knew the answer to that question, but in the meantime, can someone please help me visualize what this set looks like?  I understand that $\langle f_1,...,f_s \rangle$ is an ideal, but I don't understand its structure, if that makes sense.  In other words, I can't visualize this definition in a way that makes sense to me.  The authors do make a slightly helpful note, saying that "we can think of $\langle f_1,...,f_s \rangle$ as consisting of all 'polynomial consequences' of the equations $f_1=f_2=...=f_s=0$."
To elaborate a little more on my confusion, what I'm asking for is a less "compact" definition.  When I read this definition, for whatever reason the only thing I can come up with is $$f_1h_1+f_2h_2+...+f_sh_s.$$  But that doesn't make sense, because $\langle f_1,...,f_s \rangle$ is supposed to generate a set, not just a single polynomial.

As always, thank you all for your time.  If you find this to be a stupid or silly question, then I'm sorry to have disappointed you -- I'm a slow learner, and I get hung up on stupid things sometimes.
Oh, and Happy Halloween!
 A: You were close.

The ideal $(f_1,...,f_s)$ of $k[x_1,..., x_n]$ is the set of all polynomials which can be expressed as $h_1f_1+\cdots +h_sf_s$, for some $h_1,...,h_s \in k[x_1,..., x_n]$.

In other words, "linear combinations" of $f_1,...,f_s$ where the "linear coefficients" are arbitrary elements $h_1,...,h_s$ of $k[x_1,..., x_n]$.

Regarding the author's point about common zeros . . .

If $\bar{k}$ is an algebraically closed extension of $k$, and if there is some $a = (a_1,...,a_n) \in \left(\bar{k}\right)^n$ such that $f_1(a) = \cdots = f_s(a) = 0$, then $a$ is automatically a zero of any polynomial of the form $h_1f_1+\cdots +h_sf_s$, hence $a$ is a common zero for all members of the ideal $(f_1,...,f_s)$. 

In particular, if you've identified a common zero $a \in \left(\bar{k}\right)^n$ for $f_1,...,f_s$, then if $g \in k[x_1,..., x_n]$ is such that $g(a) \ne 0$, it follows that $g$ is not in the ideal $(f_1,...,f_s)$.

An important, nontrivial result is a partial converse: If $g \in k[x_1,...,x_n]$ is such that $g,g^2,g^3,...$ are not in the ideal $(f_1,...,f_s)$, there is some $a \in \left(\bar{k}\right)^n$ such that $a$ is a common zero of $f_1,...,f_n$, but $a$ is not a zero of $g$.
A: Perhaps a good way to understand this (or to understand anything, for that matter) is to look at examples.
The definition of ideal makes sense in any ring, so let’s look at $\Bbb Z$ first. Every ideal, you soon persuade yourself, is a set $d\Bbb Z\subset\Bbb Z$, that is, just the set of all multiples of a given number $d$. What if you try to take two numbers and look at $\langle d_1,d_2\rangle\subset\Bbb Z$? Please do this with specific numbers. Like what about $d_1=8$ and $d_2=6$? You want to describe the totality of all numbers writable in the form $8m+6n$. You rapidly see that this is an ideal in the sense of the definition, and you see that the set is equal to $2\Bbb Z$. This is basic number theory.
Now do the same thing with a ring of polynomials, but in just one variable, $R=\Bbb Q[x]$. One proves (you prove it, with Euclidean division of polynomials) that every ideal of $R$ is of form $fR$, where $f$ is a well-chosen polynomial. In fact, for a nonzero ideal $I$, you take $f$ to be a nonzero element of $I$ of least degree. So, what is $\langle f_1,f_2\rangle$, when $f_1$ and $f_2$ are two polynomials in $x$ that are given? Just as with numbers, it’s the ideal $gR$ where $g$ is the greatest common divisor of $f$ and $g$. Convince yourself that $\langle x^3-1,x^2-2x+1\rangle$ is the set of all multiples of $x-1$.
Things are no longer so simple when you have polynomials in more than one indeterminate. But at least you get some insight by looking at the simplest cases, and I hope you see that the set of all possible $h_1f_1+\cdots+h_nf_n$ is an ideal.
A: Would it help to see the see the set with explicit polynomials in place of the $f_1, f_2, \dots, f_n$?
For example, lets look at an explicit example when $n = 2$, so an ideal generated by 2 polynomials. Also, we'll work in a polynomial ring in two variables over $k$, i.e. $k[x,y]$. 
Here is the ideal generated by the polynomials $x^2 - 1$, $yx+x$.$$\langle \, x^2 - 1, \, yx+x \,  \rangle = \{ \,  f\cdot(x^2 - 1) + g\cdot(yx + x) \,  |  \,  f,g \in k[x,y] \, \}.$$
So the elements of the set are any polynomial that can be written in the form $f \cdot (x^2 - 1) + g \cdot (yx+x)$. But we can choose $f,g$ to be $\textit{any}$ polynomial we want. For example, we know $x^2 - 1$ itself is in that set because we can choose $f = 1, g = 0$. We also know that the polynomial $x^3-x+yx ^2 + x^2$ is in the set, because we can choose $f = x, g = x$.
If you are familiar with linear algebra you can maybe, in a way, draw a connection between an ideal generated by polynomials and the span of a set of vectors. You can think of it as, the set of all things that can be made from the objects defining it. 
