Why is the security level defined by $\log _{2} \inf \{ t_{i}/\varepsilon_{i} \mid i \in I \}$
The value $t/\epsilon$ is the expected amount of time required for an attacker to "win". E.g., if he has a 50% $(\epsilon = 0.5)$ probability to guess a key in $t = 10$ seconds, then the expected length of time for him to guess the key is $t/\epsilon = 10s/0.5 = 20s$

The example you provided is a bit unspecified, and it's not clear to me that $A_3$ is a "trivial modification" of $A$, what a "trivial modification" means, or why we would expect a "trivial modification" to retain the security of the original.

To be more concrete consider two examples:

First, a blockcipher $E$ with 128 bits of security (for simplicity, 128-bit keys and 128-bit blocks). Then $E'_{k1,k2}(x) = E_{k1}(E_{k2}(x))$ could be considered a "trivial modification" of $E$ and has, as you say, pretty much the same security level --- in this case, 129 bits due to a meet-in-the-middle attack.

On the other hand, if you view the blockcipher $E$ as a PRF, then the "trivial modification" $E'_{k1,k2}(x) = E_{k1}(x) \oplus E_{k2}(x)$ has significantly more security --- we've gone from ~64 bits of security (generic birthday-bound attack) to ~85 bits of security (c.f "The sum of PRPs is a secure PRF" by Stefan Lucks).

So in the absence of more specific information about your $A_3$, it's entirely reasonable that $A_3$ may have either about the same, significantly more, or even significantly less security than $A$.

A final note I don't know what text your definition comes from, but although it's a completely reasonable definition for "bits of security" --- a term often used informally --- it's a bit incomplete since it doesn't account for possible limits to the number of queries an attacker can make, or the amount of pre-computation he could perform before being provided with oracle access.