Usually the best way to start a problem like yours is to carefully expand out the definitions.
- $x\equiv y\bmod m$ means $m$ divides $x-y$,
- $y\equiv z\bmod m$ means $m$ divides $y-z$, and
- $x\equiv z\bmod m$ means $m$ divides $x-z$.
If $m=2$, for instance, these definitions say that $x-y$, $y-z$, and $x-z$ are even.
To prove the transitivity property, we need to assume that 1 and 2 are true and somehow conclude that 3 is true. 1 and 2 say that $m$ divides $x-y$ and $y-z$. We want to show that $m$ divides $x-z$. Okay. Well, whenever $m$ divides two numbers it has to divide their sum. For instance, the sum of two even numbers is always an even number. It follows that $m$ divides the number $(x-y) + (y-z)$. But $(x-y) + (y-z) = x-z$, so $m$ divides $x-z$. That is what we set out to show.
To write the proof up carefully, you would need to explain in detail why the sum of two numbers divisible by $m$ is itself divisible by $m$.
When you do enough problems like this, you start to see that there are only a few "moves" that it is possible to make. The only things that 1 and 2 give us are two numbers that are multiples of $m$ -- that's not a lot to go on. So why not see what happens when we add them together? That addition is the stroke of luck that finishes the problem.