This is a bit difficult to answer, since it’s so general, but I am sorry to hear of the difficulties so I’ll give it a try. It’ll be easier to answer these questions in the context of a problem  so, let’s make one.
A problem
Supposedly, a kid trick or treating in Montford will, on average, collect 40 pieces of candy in an hour. A random sample of 10 kids yielded only 37.2 pieces of candy with a standard deviation of 2.3. Can we say that the 40 pieces of candy is claim incorrect with
99\%
probability?
To solve this problem, we’ll need to understand the “test statistic” and the “pvalue” that you ask about. We can also compare and contrast the use of R with the use of a table.
The test statistic
The test statistic, also called the
t
score, is directly analogous to the
z
score that we used with the normal distribution. That is, if we have an observation
x
and we want to see where it lies in a population with mean
\bar{x}
and standard deviation
s
, we compute
T = \frac{x\bar{x}}{s/\sqrt{n}}.
In the example above, we’ve got
T = \frac{37.240}{3/\sqrt{10}} \approx 3.849729.
The
p
value
As always, the
p
value is the probability that we could get the observed data or worse, under the assumption of the null hypothesis. In this particular case, our null and alternative hypotheses are

H_0: \bar{x} = 40

H_A:\bar{x} \neq 40
The point behind the
t
score is this probability can be computed as the area under a
t
distribution and outside the
t
score. This looks something like so:
This is exactly what we’ve learned to compute with R using a command like:
2*pt(3.849729, 9)
# Out: 0.003907934
Thus, that shaded area (which is the same as our
p
value) is 0.003907934. We asked for a
99\%
level of confidence, so this is less than our threshold value of
0.01
and we reject the null hypothesis.
Tables
You can’t read a
p
value directly off of our class
t
table, because
t
tables are generally much more sparse than normal tables. You can read critical threshold values off of a
t
table, though, and this is enough to decide whether to reject a null hypothesis or not. If we take a look at our
t
table, we see the following:
Note that the last column corresponds to a twotail
99\%
level of confidence. If we go down to the 9 degrees of freedom row, we see that the critical
t^*
value is 3.17. Since our
t
score of 3.85 is even larger than that, we see that we must reject the null hypothesis.
Of course, if you’re working on a HW problem where the table does not include the necessary degrees of freedom, then you can use R to do the computation. On an exam, I guess I can’t go larger than 20 degrees of freedom without expanding the table.
So, this stuff is tricky  I can see why you’d call it a “difficulty spike”. I hope this helps, though, and everyone should feel free to post more questions here!