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.
My question is: Is there any disadvantages to thinking about Algebra
like this? Is there anything later in my math education that will
require me to know that I am subtracting or adding 2x to get rid of it
on this side?
As a college algebra instructor, I'm involved with remediation efforts for hundreds of students each year who have graduated high school but can't get started with college math, mostly due to incorrect concepts picked up in their prior schooling. So I would say "yes". There are some shortcuts that teachers can take to get students to pass some specific tests or programs that they are involved in; but the incorrect concepts definitely make things more difficult for students, sometimes overwhelmingly so, later on. (A majority of students that land in college remediation programs never get college degrees.)
The first thing that I would point out is that the "apply inverse operations to both sides" idea is generalizable to any mathematical operation; this allows you to cancel additions, subtractions, multiplications, divisions, exponents, radicals... even exponential, logarithmic, and trigonometric functions. (With appropriate fine print: no division by zero, square roots to both sides creates two plus-or-minus solutions, trigonometric inverses creates infinite cyclic solutions, etc.)
In contrast, the "move over and change the sign" method is not generalizable, as it only works for addend terms. This sets students on a course that requires memorizing many apparently different rules, one for each operation, which is much harder. When solving $2x = 10$, how is the multiplier of 2 canceled out? Must we remember to move it and turn it into the reciprocal 1/2? Will the students mistakenly change the sign and multiply by -1/2? Or add or multiply by -2 (I see this a lot)? How do we remove the division in $\frac{x}{2} = 5$ (probably some other rule)? How will we remember the seemingly totally different rule to solve $x^2 = 25$?
By way of analogy, I have college students who never memorized the times tables; they did manage to get through high school by repeatedly adding on their fingers, and can get through perhaps the first part of an algebra course that way. But then we start factoring and reducing radicals: "What times what gives you 54?" I might ask; "I have no idea!" will be the answer (this happened this past week; and here's a student who has effectively no chance of passing the rest of the course).
In summary: There are shortcuts or "tricks" that can get a student through a particular exam or test, which prove to be detrimental later on, as the "trick" fails in a broader context (like in this case, with any operations other than addition or subtraction). This then sets a student on a road to memorizing hundreds of little abstract rules, instead of a few simple big ideas, and at some point that complicated ad-hoc structure comes crashing down. Be polite and don't fight with your teacher to change things; but make sure to pick up a broader perspective for yourself, and share it with other students if they're willing, because you will need it later on. Take the opportunity to think about how you could improve on teaching the material, and then you may be on the path to being a master teacher yourself someday, and helping lots of people who need it.
Best Answer
The answer is already in the comments, but I'll try explain it a bit more verbosely. If you solve an equation, what you're actually doing is writing a proof that the equation is logically equivalent to some statement $x = ...$, where the dots stand for some number. You do that by writing down a series of equivalent statements, one below the other. In your case, you start out with $$ 5x + 3 = 8x + 3 $$ Then you subtract $3$ from both sides. You may do that, because two numbers are equal exactly if those numbers minus 3 are equal. So the statement $$ 5x = 8x $$ is logically equivalent to the equation you started with. Now you divide by $x$. Are you allowed to do that? Is it true that two numbers are equal exactly if they're equal if divided by the same divisor? Turns out they are, except if the divisor is zero. In that case, the division isn't allowed, and so the statement you get - even if it formally looks valid - has no more logical connection to the original equation. In particular, the original equation may have a solution, but the statement you get after the division may be simply untrue. So, by dividing by $x$, you implicitly assume that $x\neq 0$! Everything that follows will be depend on that assumption. In your case, you do the division and end up with $$ 5 = 8 \text{,} $$ an obvious contradiction. But that statement is only equivalent to the original equation if $x \neq 0$ - after all, we had to assume that to get this far. So we now know that, indeed, $x$ has to be zero for the equation to hold, since assuming that it isn't zero got us into trouble. We don't yet know if it actually will hold for $x=0$, thought, so the last step is to set $x=0$ in the original equation, and verify that it checks out.