When we call grow in the push method, if we grow the array by one (or few) elements, then the number of times we call "grow" linearly increases with the number of times we call "push."

The function grow itself is a linear-time operation. So we have a situation that looks like $O(n) \times O(n)$, which gives us $O(n^2)$ quadratic runtime for $n$ "push" operations.

Another way to see this is that for one million push, the (computational) work performed by the "grow" operation (in total) will be as follows.

$$ 1 + 2 + \dots + 999999 = 499999500000 \approxeq \text{1 billion} $$

The above shows the $O(n^2)$ total runtime for $n$ "push" operations. The amortized cost of push will then be $\frac{O(n^2)}{n}=O(n)$.

If we grow the array by doubling its size, the number of times we call grow logarithmically increases with the number of pushes.

Let's say we start with an array of 1 element, and then we do $n$ push. The total work done is as follows.

$$ 1 + 2 + 4 + 8 + \dots + 2^{\lg n} = 2^{(\lg n) + 1} - 1 $$

The total above is calculated by adding up all the powers of two.

Note that $2^{\lg n} = n$ (recall $\lg n$ is $\log_{2} n$. Moreover, look at rule #7 of logarithm rules).

So we have the following:

$$ 2^{(\lg n) + 1} - 1 = \left ( 2^{(\lg n)} \times 2 \right ) - 1 = 2n - 1 \in O(n) $$

Thus the total runtime is $O(n)$ for $n$ "push" operations. The amortized cost of push will then be $\frac{O(n)}{n}=O(1)$.