Understanding disk requirement to make fail over cluster Please take a look at the following screenshot.  



I am trying to make SQL Server 2012 Fail Over Cluster using two nodes. As you know we need to have a shared SAN which will be used by both nodes to store data but I have some confusion regarding it.

I was under the impression that we need one drive for e.g. Z for "Temp DB directory" which will be shared by both nodes to store data.

But someone told me that these will be two drives actually for each node and for these nodes it will appear as one. So I am not sure what is true. May be we are both saying same thing!?
That all depends on which configuration you're seeking. As @EricHiggins noted, a traditional cluster will have shared resources, including disk, network name(s) and IP(s), and the service itself. SQL Server is a cluster-aware application and will interact with the Windows cluster service. When a failure (or manual failover) is detected by the cluster service, the SQL Server service is stopped (taken offline) and the instance resources (name, IP, disk, and other cluster resources) are migrated/transferred to the another node in the cluster. When the resources are fully transferred, the SQL Server service is started and the instance is available.

With SQL Server 2012 Enterprise Edition, the concept of Availability Groups was introduced to the system. Availability Groups are a shared-nothing environment that uses failover clustering for network naming and listening, but uses mirroring to protect at the database level. 

There are many more details to go into to draw out the distinction between (and the benefits of) each of these solutions. However, to directly answer your question, if you're going to cluster using failover clustering then the TempDB LUN would be shared. If you're going with Availability Groups, then you'll double your storage, but be able to protect at the database level (instead of the instance level). 

I'd need more information about "appears as one" statement - to what, or whom, does it "appear as one?"