First, for a bit of clarifying terminology: rather than proving an amortized insertion cost of $O(\lg n)$ and an amortized deletion cost of $O(1)$, you are using those amortized costs to prove something about the total cost of a sequence of insertions and deletions, starting with an empty heap. This might be a subtle distinction, but it is fairly important (at least to me). If you allow yourself to start from a non-empty heap, you would also need to allow yourself to start with a $\Theta(n \lg n)$ potential, or else the deletion amortized cost won't work. Or, if you were to add in an operation for the worst-case linear time build-heap operation, you would also need for that linear time operation to add $\Theta(n \lg n)$ to the potential function.
To use your desired amortized runtimes, each insertion has to prepay for a deletion to come later. So, for instance, when inserting into a heap of size $n$, you can pay for the actual insertion $O(\lg n)$, but your amortized cost will pay that plus an additional $O(\lg n)$ for the worst-case runtime of deleting from a heap of size $n+1$. Now, every time you delete from a non-empty heap? That deletion's actual cost has already been prepayed by a previous insertion's amortized cost. It costs 0 (amortized) to delete. The only deletions that you still need to pay for are the ones from an empty heap, for when your deletions outnumber prior insertions, and those cost $O(1)$.
While these bounds are fine to use for a proof of efficiency for a sequence of insertions and deletions, they are slightly odd, in that in real life, the deletions generally take expected $\Theta(\lg n)$ time, and insertions take expected $\Theta(1)$ time, for some reasonable definitions of operating on a random heap of size $n$.
While I have presented this in terms of "prepaying" (or the accounting method), in this case, just writing it down as a more formal equation directly transforms it into the potential function method. You will be able to prove that the potential of a heap with $n$ elements is at least $\sum_{i=1}^n f(i)$ where $f(i)$ is the worst-case price to delete the min item from a heap if $i$ elements. You already start your question with a given $O(\lg i)$ bound on $f(i)$.
