What form does a Taylor series general term need to be in? Why don't my general term equation give me the first term sometimes? So this is written in my book for Taylor series:

Most importantly, I notice that the exponent after $(x-a)$, and the factorial, and the n that currently being iterated over are all equal.
To demonstrate my confusion, let's look at the Taylor series for $ln(1+x)$ and $xe^{-2x}$.
so:
$$f'(x) = \frac{1}{x+1},\quad f''(x) = \frac{-1}{(x+1)^2},\quad
f'''(x) = \frac{2}{(x+1)^3},\quad f^{(4)}(x) = \frac{-6}{(x+1)^4}$$
and
$$f'(0) = 1, \ f''(0) = -2,\ f'''(0) = 2, \ 
f^{(4)}(0) = -6
$$
so notice the general term starts at $n = 1$ but the exponent to x is only n. Is this a problem?
so $$
\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} x^n
= x - \frac{x^2}{2} + \frac{x^3}{3}$$
So the general term here seems to accurate represent the Taylor series even though the n term doesn't match up with the exponent term for x. 
However, $xe^{-2x}$ is different:
so finding the Taylor series form:
we know that $e^x$ is: $\sum_{n=0}^{\infty} \frac{x^n}{n!}$. 
So
$$e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!}, \
e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^nx^n}{n!},\ 
xe^{-2x} = \sum_{n=0}^{\infty} \frac{(-2)^nx^{n+1}}{n!}$$
So according to the formula, the first term is just f(a) which should equal the general formula when I plug in n = 0 right? But theres a mismatch. When I plug in n = 0, I get $x$. But f(0) = 0. So what gives? What am I doing wrong here?
 A: Let me repeat your question in a concise way. 

Let $f(x)=\ln (1+x)$ and $a=0$. On the one hand, formula (6) you quoted form the book says that the $0$-th term of the Taylor series should be $f(a)=f(0)=\ln(1+0)=0$. On the other hand, one has
  $$
\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} x^n
= x - \frac{x^2}{2} + \frac{x^3}{3}+\cdots\tag{1}
$$
  The term $x$ in (1) does not match the $0$-th term in the formula (6), what is going wrong?

It is not that there is mismatch but that you match the formulas in a wrong way. The equality in (1) could be written as
$$
\ln(1+x) = \color{blue}{ 0 +} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} x^n
= \color{blue} {0+}x - \frac{x^2}{2} + \frac{x^3}{3}+\cdots\tag{2}
$$
And (2) "matches" formula (6):
$$
f(0)=0\;,\  f'(0)=(-1)^{\color{red}{1}-1}(\color{red}{1}-1)!=1\;,\\
f''(0)=(-1)^{\color{red}{2}-1}(\color{red}{2}-1)!=-1\;,\\ 
f'''(0)=(-1)^{\color{red}{3}-1}(\color{red}{3}-1)!=2!\;,\\
\vdots
$$

Now, let us look at another Taylor series you mentioned in your post:
$$
g(x)=xe^{-2x} = \sum_{n=0}^{\infty} \frac{(-2)^nx^{n+1}}{n!}\tag{3}
$$
Again, what confuses you is that $g(0)=0$, which should be the $0$-th term in the Taylor series, but for $n=0$, the term in (3) is $x$. 
The problem is that the $n$ in (3) is NOT the same as the $n$ in (6)!! 

In general, if
$$
\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty b_n,
$$ 
one can not conclude that $a_n=b_n$. 

Exercise. Find the Taylor series for the function $f(x)=x^3$ at $a=0$ using formula (6) and see how your result "matches" the "general formula".
