What is written below is intended to help a bit with the intuition. All series discussed below have non-negative terms.
Call a series $\sum x_n$ bad if it diverges, and good if it converges.
Comparison Test 1 says that if $\sum b_n$ is good, and $0\le a_n\le b_n$ for all $n$, then $\sum a_n$ is good.
In cruder terms, if $\sum b_n$ doesn't blow up, and the $a_n$ are smaller, then $\sum a_n$ doesn't blow up.
Now suppose $\sum a_n$ is bad, and $a_n\le b_n$ for all $n$. Then $\sum b_n$ cannot be good. For if it were, then by Comparison Test 1, $\sum a_n$ would be good. But it isn't. So we have obtained:
Comparison Test 2: If $\sum a_n$ is bad, and $0\le a_n\le b_n$ for all $n$, then $\sum b_n$ is bad. If $\sum a_n$
blows up, and the $b_n$ are bigger than the $a_n$, then $\sum b_n$ blows up.
There are useful variants of the comparison tests. For one thing, what happens for "small" $n$ doesn't matter. If $a_n\le b_n$ for large enough $n$, and $\sum b_n$ is good, then $\sum a_n$ is good. Also, if $\sum b_n$ is good, and there is a positive constant $c$ such that $0\le a_n\le cb_n$, then $\sum a_n$ is good.
In order to apply the Comparison Tests successfully, we need to have a basic stock of series whose convergence/divergence we know about.
That basic stock need not be large. In practice, most comparisons are done with geometric series, or with $p$-series $\sum\frac{1}{n^p}$.
We do have to be careful about the direction of the inequalities. Suppose we know already that $\sum \frac{1}{n^2}$ is good. That doesn't help at all with $\sum \frac{1}{n}$. For we have $\frac{1}{n^2}\le \frac{1}{n}$. So the terms of $\sum \frac{1}{n}$ are bigger than the terms of a good series. Not useful: A series bigger than a good series can easily blow up.
Very informally, a series of positive terms converges (is good) if the terms approach $0$ "fast enough." It turns out that $\frac{1}{n^2}$ approaches $0$ fast enough, but $\frac{1}{n}$ does not approach $0$ fast enough.
Interestingly, $\frac{1}{n}$ approaches $0$ almost fast enough. For example, $\frac{1}{n^{1.01}}$, which approaches $0$ only a little faster than $\frac{1}{n}$, is "fast enough."