How Blockchain ensures transactions are not lost? I have a doubt about how the Blockchain system avoids transactions to be lost.

Let me use an easy example: 

Let's say there are only 2 mining pools (A and B), which are both trying to find a hash for block 5 to add it to the chain. 

Now, A's block 5 can differ from B's block 5 in terms of transactions. E.g. the transaction "Alice buys a coffee for 5 BTC" can be in the block 5 for pool A, but not in block 5 for pool B (e.g. because this transaction arrived later to the node B compared to the node A).

Now, if the winning node for this block is the node B, the transaction "Alice...." won't be stamped in the ledger. 

When the competition for the block 5 is over, are the "losing" pools/nodes (in this case the pool A) dropping all the transactions in the block they were working on? and take new transactions from the memory pool?

In this case, if the pools A and B start looking for block 6's hash, the transaction "Alice...." could be in block 6 for pool B, but it WON'T be in block 6 for pool A (pool A dropped this transaction when loosing the "race" for hashing block 5).

What if node A hashes block 6 before B? Now B will drop the transaction in its block 6, so the transaction "Alice...." is disappeared definitely.

Please correct me

Cheers
Why do you think that the transaction will be "lost"?

It will simply stay in the mempool as long as it is broadcasted by the sender and it will stay there until a miner decides to work with it, and thus confirm it in a new block.

A miner dropping the transaction because he doesn't want to include it in the next block that they're mining simply won't make it disappear.

If node B fails to get the next block, where they intended to confirm Alice her transaction, it would simply mean that her transaction will be in the mempool until another miner uses it in his block.

Note that this can happen simultaneously, it seems to me that you think that it is only possible for a transaction to be mined by 1 miner at the time, although this isn't the case.