Question 1: This is resolved by precise use of the word "problem". You seem to be using it in a different way from how your book defines it (and how it's usually defined). $s$ is a problem instance. For example, the traveling salesman problem isn't just one set of distances, but the entire class of problem instances, where each particlar problem instance is specified by a particular set of distances. So $s$ is a string specifying a particular instance of the problem $X$.
Question 2: $|s|$ is the length of the string $|s|$ and $p$ is a polynomial function, so $p(|s|)$ is just that polynomial function applied to the length of $s$. Similarly, $|t| \le p(|s|)$ means that the length of the string $t$ is bounded by the polynomial function $p$ of the length of the string $s$. In other words, the length of the description of the certificate is at most polynomial in the length of the description of the problem instance.
[Edit in response to the comment:] Going through your additional questions one by one:
"But how are the amount of work done by B based on the size of those strings?" I don't understand that question; it's ungrammatical.
"I see time complexity as something based on a problem's input size, but I have trouble seeing the size of those strings as the input size." What does "those strings" refer to here?
"Another question: In the "sāX iff..." statement, is the X here refering to a NP problem?" At that point it isn't, since that statement is in the definition of an efficient verifier and there's no talk about NP yet. But the subsequent definition of NP in terms of efficient verifiers means that if there is such a $B$, then $X$ is by definition an NP problem.
"Why does B(s,t) have to be yes? Is a problem considered NP if a solution can be verified to be false in polynomial time?" The term "solution" isn't used in the definition, but I'm guessing that you might mean something like this: Consider the decision version of the traveling salesman problem, i.e. "Is there a tour of length $
$B$ is a polynomial-time algorithm. The definition doesn't say that it runs in polynomial time only if the result is "yes"; it always runs in polynomial time. In particular, it runs in polynomial time and answers "no" (or anything else other than "yes") if you feed it a "false solution".
You're right more generally, however, in noticing that there's an asymmetry here. But the asymmetry is not in the ability to check certificates, but in their existence. Whereas NP is defined as the class of problems for which there exists a (polynomial-time-verifiable) "yes" certificate for all "yes" instances, co-NP is defined as the class of problems for which there exists a (polynomial-time-verifiable) "no" certificate for all "no" instances. In the case of the decision version of TSP, such a "no" certificate would not be a "false solution" in the sense of a tour with length $\ge L$ (which proves nothing either way), but a (polynomial-time-verifiable) proof that no tour of length $the Wikipedia article on co-NP.