Learning to differentiate between the sum and product rules I'll try and explain the difference the best way I can. Please, tell me if I make sense.

Calvin wants to go to Milwaukee. He can choose from $3$ bus services or $2$ train services to head from home to Chicago. From there, he can choose from $2$ bus services or $3$ train services to head to Milwaukee. How many ways are there for Calvin to get to Milwaukee?

There are two ways I look at it:
Indpendence:


*

*if he chooses a bus service, then that affects his choice of a train service. If he chooses a bus, he can’t choose a train. So add them: $(1)$  $3$ bus services + $2$ trains $(2)$  $2$ bus services + $3$ trains

*choosing a means of transport in Chicago doesn’t depend on what he chose in his hometown. So multiply # of ways to get from his town to Chicago with the number of ways to get to Milwakee: $5^2$.


Mixture/Non-mixture(I am sure there's much better way to put it):


*

*If he adds the number of ways to get to Chicago to those that let him get to Milwaukee, he’ll get: $\{BS_1, BS_2, BS_3, BS_4, BS_5, TS_1, TS_2, TS_3, TS_4, TS_5\}$ where $BS$ stands for Bus Service and $TS$ stands for Train Service. To get from his house to Milwaukee he would need sets like  $\{BS_1, TS_2\}, \{TS_1, BS_2\}, \{TS_1, TS_3\} \ldots 25$ in total. To get such a mixture he would always have to multiply 



Suppose the group of twelve consists of five men and seven women. How many five-person teams contain at most one man?

Answer: $\binom 51 \binom 74 + \binom 52 \binom 73 + \binom 53 \binom 72 + \binom 54 \binom 71 + \binom 55 \binom 70.$
Independence:


*

*We sum the number of these subsets because they are not independent.  For example, if we choose  teams with one man and four women we can’t choose  teams with two men and three women.

*To get the number of teams  with one man and four women we use the product rule because however we choose the men isn’t going to affect our choice of women. 


Mixture/Non-mixture:


*

*Consider $\binom 51 \binom 74.$ Why don’t we add them? If we add them we get a set of $\{m_1\}, \{m_2\},\{m_3\}, \{m_4\}, \{m_5\}, \{w_1, w_2, w_3, w_4\}, \{w_7, w_1, w_3, w_5\}, \{w_3, w_2, w_6, w_3\}$ etc where $m_i$ is the $i$th man and $w_i$ is the $i$th woman. There are total of thirty five 4-sets. On the other hand we need sets mixed like $\{w_1, w_7, w_3, m_1, w_2\}, \{w_1, m_3, w_2, w_6, w_5\}$ etc. To achieve such mixture we need to multiply.

 A: When counting ways to perform tasks with an "and" connective, that is simply an conjunction of independent tasks, also often known as "performing in series", we count the product of ways to perform each subtask.
If Bob catches one of two services, and then one of three service, when the choice of the first service does not influence the choice of the next, nor conversely so, then he is performing a two choices in series.   We multiply to count: $2\times 3$.

When counting ways to perform tasks with an "or" connective, that is where the options available are dependent on some critical choice, that is a disjunction of alternatives, less often known as "performing in parallel", we count the sum of ways to perform each alternative subtask.
If Bob can take one path or another, where the choice of which influences the availability of services (and more importantly, how many services are on offer), then we must add the counts of paths available from each choice.
Note: When the count of services available is not affected by the choice of paths, then using the "sum rule" is equivalent to the "product rule", because $3+3 = 2\times 3$, and so on.   It is only where we would not be adding a series of identical counts that we (obviously) cannot multiply.

Now, let's look at Calvin's task.

Calvin wants to go to Milwaukee.   He can choose from $3$ bus services or $2$ train services to head from home to Chicago.   From there, he can choose from $2$ bus services or $3$ train services to head to Milwaukee. How many ways are there for Calvin to get to Milwaukee?

It is $(3+2)\times(2+3)$, do you see why?


Suppose the group of twelve consists of five men and seven women. How many five-person teams contain at most one man?

The number of women we can select depends on how many men we select.  So the task is to select: one man and four women, or two men and three women, or three men and two women, or four men and one woman, or five men and no women.
$$\binom{5}{1}{\times}\binom{7}{4}~+~\binom{5}{2}{\times}\binom{7}{3}~+~\binom{5}{3}{\times}\binom{7}{2}~+~\binom{5}{4}{\times}\binom{7}{1}~+~\binom{5}{5}{\times}\binom{7}{0} \\ = \binom{12}5-\binom 75$$
tl;dr If you can say "do this or that" you have an addition.   If you can say "do this and (then) that" you have a multiplication. 
If you say "do this but not that" you have a subtraction.   As in "select any five from twelve people but not all five from the seven women."
