Found a peculiar solution I though I would share:

def appendAndDelete(s, t, k):
    i = 0
    
    # Comment 1
    
    for j in range(len(min(s,t))):
        if s[j] == t[j]:
            i += 1
        else:
            break
            
    # Comment 2
    
    if k >= len(t) + len(s):
        return 'Yes'
     
    # Comment 3
    
    else:
        min_ops = len(t) + len(s) - 2*i
        if min_ops <= k and (k - min_ops)%2 == 0:
            return 'Yes'
        return 'No'

Comment 1

Calculate the length of the substing of s (starting at the beggining of s) which we do not need to change because it is the same as the corresponding substring of t.

Comment 2

If k >= len(t) + len(s) the answer is always yes because we can delete all the letters of s, continue deleting the empty string if k > len(t) + len(s) and then add the elements of t

Comment 3

min_ops = len(t) + len(s) - 2*i calculates the minimum number of operations needed to transform s to t. If min_ops > k the answer is no. If min_ops = k the answer is yes. If min_ops < k, the answer is yes if the remaining operations are an even number. In this case, after transforming s we would use the remaining number of operations to add and delete an item (or vice versa) until we use all of k.