You don't assume P(12); you have toproveit. Also, it helps to write explicitly what P(n) is. This helps to avoid confusion later in the proof, not to say that all symbols must be defined prior to their use.

Again, it helps to write what exactly the induction hypothesis is, which is the real question here.

You can assume the first equality as long as k > 3 (I am assuming that in this particular equality all operations are on natural numbers), which is true since k >= 12 by assumption. As for P(k - 3), i.e., the fact that there exist nonnegative integers x and y such that k - 3 = 4x + 5y, you cannot assume this for all k >= 12. Indeed, you have shown P(12 - 3) and P(13 - 3), but not P(14 - 3). If you leave it like this, you have a gap in the proof for P(15), P(19), P(23), etc. To cover this gap, it is sufficient to prove separately P(15), i.e., P(k + 1) when k = 14.

Suppose we are only asked to prove P(k) for k >= 12. Then one has to prove four base cases: P(12), P(13), P(14), P(15) together with the inductive step: P(k - 3) implies P(k + 1) for all k >= 15 (which is the same thing as P(k - 4) implies P(k) for all k >= 16). This way, the base case P(12) plus the inductive step will cover P(12), P(16), P(20), etc.; the base case P(13) will cover P(13), P(17), P(21), etc.; the base case P(14) will cover P(14), P(18), P(22), etc.; and finally P(15) will cover P(15), P(19), P(23), etc. Together, they cover all integers >= 12. For each number n >= 12, you can think back to the chain of numbers m such that P(m) is involved in the proof of P(n): here it is n - 4, n - 8, etc. For each of those numbers m, P(m) must be proved either in the inductive step or in the base case.

There is also a way to prove this statement by only going from P(k) to P(k + 1). Note that 4(-1) + 5 = 1, so, if 4x + 5y = k, then 4(x - 1) + 5(y + 1) = k + 1. The problem is when x = 0, i.e., 5y = k. But then y > 3 since k >= 12 and 4(4) + 5(y - 3) = k + 1.