When am I allowed to do this:
$$\sum_{n=1}^{\infty} f_n + g_n = \sum_{n=1}^{\infty} f_n + \sum_{n=1}^{\infty} g_n $$?
I know I can do it if both $\sum_{n=1}^{\infty} f_n$ and $\sum_{n=1}^{\infty} g_n$ are convergent, but can I do it if one of them is divergent and the other convergent?
If one is convergent and one divergent the sum will diverge. The place you have trouble is that both the sum of $f_n$ and the sum of $g_n$ can be divergent while the sum is nicely convergent. For example $$f_n=2^n\\g_n=-2^n\\f_n+g_n=0$$
If any 2 sums exist, then the third exists and the equality holds. e.g. suppose $f_n+g_n$ and $f_n$ is summable. Then $-f_n$ is summable, and apply the known result for $F_n := f_n + g_n$ and $G_n = -f_n$ to see that $F_n + G_n$ is summable, with the equality.
It is even worse. Consider two divergent series: $\sum_{n} f_n = \sum_{n} \frac1n$ and $\sum_{n} g_n = - M \sum_{n} \frac1{M n}$ with some natural number $M$. They both diverge, since what we have here are both harmonic series. Now add the two. What you get is
$$ \sum_{n} f_n + g_n = \sum_{n} \frac1n - M \sum_{n}\frac1{M n} \\ = 1 + \frac12 + \cdots + \frac{1}{M-1}+ \frac{1-M}{M} + \frac{1}{M+1} + \cdots + \frac{1}{2M-1}+ \frac{1-M}{2M} + \frac{1}{2M+1} + \cdots \\ = \ln(M) $$ So $\sum_{n} f_n + g_n $ is convergent. The last result may be surprising (had you expected it is zero?) but can be found here. A prominent special case is $M=2$ which is $$ \sum_{n} f_n + g_n = \sum_{n} \frac1n - 2 \sum_{n}\frac1{2 n} \\ = 1 - \frac12 + \frac13 - \frac14 + \cdots = \sum_{n} \frac{(-1)^{n}}{n} = \ln(2) $$ which is the well known alternating harmonic series.
