Algorithm in Bottom up approach
if a[n]== null
a[n] = fibo(n-1) + fibo(n-2)
There are a couple of quick things to mention before answering your question about the time complexity. The reason for this is that the time complexity at least partially depends on these answers.
First, there seems to be a bug in your program as you have an array 'a' for the base conditions (Fibbinacci numbers 0 and 1) and some array 'm' which is set in the fibo function, but never used again. More importantly, when you reach n=1 or n=0, you return the value of m[n] which is entirely unknown. So, I'm going to assume the algorithm is rewritten as follows:
a=0,a=1 integer fibo(n) if a[n]== null a[n] = fibo(n-1) + fibo(n-2) return a[n]
Okay, second problem. Let's assume that that a is always defined as at least n+1 integers. There needs to be enough room for the incoming data. This is important because c++ will let you overwrite values at the n+1th index. It's out of bounds and wrong, but c++ doesn't give those sorts of protections. It is up to you as the programmer to verify boundary conditions like that. (I'm assuming c++ because this is tagged with c++. The code looks more like python, which has its wrap-around indices which are problematic on their own.)
Third, let's assume that you don't start with a new array 'a' for each run of the algorithm. This is important because if a stores already-calculated values then you will save time on calculation by not having to re-evaluate those values. That time savings is a great thing even if it won't affect how I calculate time complexity.
Great. Let's get started with your question. Let's use the image below to answer it. When you start the algorithm at n you are going to make two recursive calls for fibo(n-1) and fibo(n-2) BUT they do not happen simultaneously. Instead the first call for fibo(n-1) takes place and must be 100% complete before the second call for fibo(n-2) begins. That call is represented by the green line from n-1 on the nth line to the n-1th line.
Now, those green lines apply to each recursion down the line until you reach the fibo(1) call. That call terminates early because a[n] is NOT null. Finally the second call for fibo(0) is executed and it also terminates early because a[n] is not null. Okay, so much for the first set of recursive calls.
As each recursive call returns, the second call (represented by the orange broken line) is made, but a[n] is no longer null, so that call terminates early and the call returns up to the next layer.
So, let's count the number of calls. From n to 1 is n-1 recursive calls. At the end there is one additional call to fibo(0) so that is n recursive calls. Then on the way up there are n-2 additional calls which terminate early. So, altogether we have 2n-2 calls which is O(n).
Of course, if you call fibo(k) and then fibo(k+x) you will only need to do the first 2x calls because everything from fibo(k) down is already known. It is a considerable savings after the initial investment. Any questions?
Regarding O(2n)=O(n), that is a good follow up. Big-O complexity rules say that we are interested in the order-of-magnitude when you compare efficiency. So, suppose that you were looking at a n=1000. O(n)=1000, O(2n)=2000, but O(n2)=1,000,000. O(n) is more or less the same as O(2n), but if you compare them with O(n2), that is a huge difference. Similarly, if you have O(n+1)=1001 that isn't much different from O(n). So, in general we say that the leading term, the most important value in the equation is what is important. We aren't really interested in extra terms. We aren't really interested in specific coefficients because they don't really affect the outcome.
If you still have questions, see this site for some additional information. https://justin.abrah.ms/computer-science/big-o-notation-explained.html