The function $\frac{x^3}{3}+2x^2$ does not grow exponentially faster than $x^2+4x$. The function $\frac{x^3}{3}+2x^2$ is a polynomial, and it grows roughly like $(x^2+4x)^{3/2}$. That is not dramatically faster than $x^2+4x$. The following is an example of faster relative growth that is more familiar.
Look at the area of the region under the curve $y=x$, above the $x$-axis, from $x=0$ to $x=w$. If you draw the "curve" $y=x$, you will see that the region is a triangle with base $w$ and height $w$, so the region has area $w^2/2$. (The same result can be obtained by integration, but that's overkill.)
So the function grows like $w$, and the area grows like $w^2$, more precisely like a constant times $w^2$.
The most "extreme" example is a constant function like $f(x)=1$. The area from $0$ to $w$ under the curve is $w$. This grows much faster than $f$, which does not grow at all, but the rate of growth of the area would not be considered fast. And the fact that the area of a rectangle of constant height grows linearly with the base is unsurprising.
In general, imagine a curve $y=f(x)$, where $f(x)$ is positive, and, for simplicity, increasing. Let $W$ be the region below $y=f(x)$, above the $x$-axis, from $x=0$ to $x=w$. Then the region $W$ is contained in the rectangle with base the interval from $0$ to $w$, and height $f(w)$. So the area of $W$ is $\le wf(w)$. Thus, if $f(w)$ grows "fast," then the rate of growth of the area is not much faster than the rate of growth of $f(w)$.
I will not explain calculus. There are many websites that do it very well, probably much better than I can, so I'll leave it to them. I will address prerequisites for calculus with an emphasis on procedure rather than deep understanding. At some point, you would do well to revisit these concepts in a slow, measured pace, being very careful and filling in all the details. However, your position seems more of an "in over my head" thing so I'll try to address it from that perspective.
There are only two big concepts in calculus, the derivitive (verb: to differentiate) and the integral (verb: to integrate). Both of these are special uses of the extremely broad concept of a limit, but many introductory calculus classes only touch on limits in a very superficial way. Certainly to do a first physics course with calculus you will not need a deep understanding of limits, only an appreciation for why they let us get the results we want.
If you do not have a good grasp on trigonometry, you can still do/understand calculus but it will seem rather artificial. You should know the sine and cosine functions and it will help to know the other four that usually accompany them. You should know the unit circle and the special values on it. Some basic identities will make your life easier.
For limits, you can get by with an intuitive understanding of functions, and of real numbers. It is more helpful to have a good intuition for rational numbers, for example you should know that there are infinitely many rationals in between any two real numbers. You must understand the notion of the domain of a function. Experience with rational functions is extremely valuable. To do calculations with limits, you should be very comfortable with simplifying rational expressions, including domain issues and extraneous solutions. Again, you can do without rational functions, but you are much better equipped to understand the significance of limits if you can manipulate them without much trouble.
For derivatives, you will need to be familiar with operations of functions: addition, subtraction, multiplication, division. A special emphasis on composition of functions: for you this will probably be the most important prerequisite for solving physics problems. You should be familiar with but do not need to be extremely comfortable with implicitly defined curves: for example, the circle is not given in $y=f(x)$ form, but it is still well-defined. You should understand the domain issues that can arise when converting implicitly defined curves into function form.
For integrals, there are a lot of skills you could need, but I will try and keep it to a bare minimum. I would not try to understand the real definition of an integral (your resources may call it a Riemann integral), but it is absolutely essential that you understand the intuition behind it, and its link to limits.
Of course you must be familiar with finding area of basic shapes. You must be extremely comfortable with reading $\Sigma$ notation; if not in reading it directly, then at least to translating it into $+$ notation (you do not need to be able to write $\Sigma$ notation well). However, the most useful skills for cracking integrals are pattern-recognition and persistence; they can sometimes require quite a bit of creativity to solve.
There is another (shorter) list which I think is equally important for your situation: things which you are not expected to know, but are expected to pick up during a calculus class. These include deeply understanding inverse functions, familiarity with theorems of the form "If … then there exists …", high comfort with implicit curves, high comfort with recognizing compositions of functions [you will need to pick this one up], the significance of variables as distinct from numbers, skill at visualizing 3D space, distinction between functions and their values at points.
At some point while learning derivatives, you will come across the notion of related rates. Please learn this very carefully. Many students struggle a lot with this section — including me — but it is a very important use of calculus for physics. Perhaps it will not come up directly in your class, but if you know it well you will see it hiding just behind the things you discuss.
Best Answer
Suppose $f(x)=x^3$.
Then $f(2)=8$ and $f'(x)=3x^2$, so $f'(2)=3\cdot2^2=12$.
That means when $x=2$ and $f(x)=8$, then $f(x)$ is changing $12$ times as fast as $x$ is changing.
So suppose $x$ goes from $2$ to $2.0001$, the change being $\Delta x=0.001$. Then $f(x)$ should go from $8$ to about $8.0012$, the change being about $\Delta f(x)=0.0012$, i.e. $12$ times as much. Why not exactly $12$ times as much? Because as $x$ changes from $2$ to $2.0001$, the derivative, thus the rate of change, does not remain exactly $12$.
(In fact, $f(2.0001)=2.0001^2 = 8.0012\ 00060001$, so $12$ times as much is pretty close.)
When calculus is taught to liberal arts majors, this kind of thing should be considered far more important than chanting "n x to the n minus one", which is what is typically taught. A standard calculus course for math majors was created, then watered down to get a calculus course for English majors, then very large numbers of the latter were encouraged to take calculus and told it would look good on their resumes. The learn to answer questions like "Find the derivative of $f(x)=\sec^3(5x+2)$" without finding out that differential calculus is about instantaneous rates of change or why it is important in the development of science and engineering over the past few centuries. Mathematicians feel forced to go along with the system, which must be maintained because those students bring in tuition money. Mathematicians who sit on curriculum committees in large departments are not the ones who are assigned the task of teaching first-semester calculus, and don't know what goes on there. The ones who do know are often less experienced and are not the ones who will develop alternative sorts of courses, and must devote their energies to publishing research so that they can keep their jobs. If you try to include things like this in a calculus course at the expense of chanting "n x to the n minus one", the sort of student who's there only to get a grade says "Will this be on the department's common final exam in this course? No? Then why are you wasting our time on it? My father donates a lot of money to this university and he will complain to the Dean about you." ONLY mathematicians can change this situation, so they cannot forever plead that they were only following orders.