The cross-entropy error function in neural networks
One way to interpret cross-entropy is to see it as a (minus) log-likelihood for the data $y_i'$, under a model $y_i$.

Namely, suppose that you have some fixed model (a.k.a. "hypothesis"), which predicts for $n$ classes $\{1,2,\dots, n\}$ their hypothetical occurrence probabilities $y_1, y_2,\dots, y_n$. Suppose that you now observe (in reality) $k_1$ instances of class $1$, $k_2$ instances of class $2$, $k_n$ instances of class $n$, etc. According to your model the likelihood of this happening is:
$$
P[data|model] := y_1^{k_1}y_2^{k_2}\dots y_n^{k_n}.
$$
Taking the logarithm and changing the sign:
$$
-\log P[data|model] = -k_1\log y_1 -k_2\log y_2 - \dots -k_n\log y_n = -\sum_i k_i \log y_i
$$
If you now divide the right-hand sum by the number of observations $N = k_1+k_2+\dots+k_n$, and denote the empirical probabilities as $y_i'=k_i/N$, you'll get the cross-entropy:
$$
-\frac{1}{N} \log P[data|model] = -\frac{1}{N}\sum_i k_i \log y_i = -\sum_i y_i'\log y_i =: H(y', y)
$$

Furthermore, the log-likelihood of a dataset given a model can be interpreted as a measure of "encoding length" - the number of bits you expect to spend to encode this information if your encoding scheme would be based on your hypothesis. 

This follows from the observation that an independent event with probability $y_i$ requires at least $-\log_2 y_i$ bits to encode it (assuming efficient coding), and consequently the expression 
$$-\sum_i y_i'\log_2 y_i,$$
is literally the expected length of the encoding, where the encoding lengths for the events are computed using the "hypothesized" distribution, while the expectation is taken over the actual one.

Finally, instead of saying "measure of expected encoding length" I really like to use the informal term "measure of surprise". If you need a lot of bits to encode an expected event from a distribution, the distribution is "really surprising" for you.

With those intuitions in mind, the answers to your questions can be seen as follows:


Question 1. Yes. It is a problem whenever the corresponding $y_i'$ is nonzero at the same time. It corresponds to the situation where your model believes that some class has zero probability of occurrence, and yet the class pops up in reality. As a result, the "surprise" of your model is infinitely great: your model did not account for that event and now needs infinitely many bits to encode it. That is why you get infinity as your cross-entropy. 

To avoid this problem you need to make sure that your model does not make rash assumptions about something being impossible while it can happen. In reality, people tend to use sigmoid or "softmax" functions as their hypothesis models, which are conservative enough to leave at least some chance for every option.

If you use some other hypothesis model, it is up to you to regularize (aka "smooth") it so that it would not hypothesize zeros where it should not.
Question 2. In this formula, one usually assumes $y_i'$ to be either $0$ or $1$, while $y_i$ is the model's probability hypothesis for the corresponding input. If you look closely, you will see that it is simply a $-\log P[data|model]$ for binary data, an equivalent of the second equation in this answer.

Hence, strictly speaking, although it is still a log-likelihood, this is not syntactically equivalent to cross-entropy. What some people mean when referring to such an expression as cross-entropy is that it is, in fact, a sum over binary cross-entropies for individual points in the dataset:
$$
\sum_i H(y_i', y_i),
$$
where $y_i'$ and $y_i$ have to be interpreted as the corresponding binary distributions $(y_i', 1-y_i')$ and $(y_i, 1-y_i)$.