Question: I'm inclined to make this claim because the functions epigraph is $\{(x,t) : t \ge f(x)\}$. But to be a convex cone, it must be closed under the usual

$$\theta\_1 (x\_1,t\_1) + \theta\_2 (x\_2,t\_2)$$

for $\sum \theta = 1$ and $\theta\_i \in [0,1]$. But that then implies:

$$\theta\_1 t\_1 + \theta\_2 t\_2 \ge f(\theta\_1 x\_1 + \theta\_2 x\_2) $$

which is the same as

$$f(\theta\_1 x\_1 + \theta\_2 x\_2) \le \theta\_1 f(x\_1) + \theta\_2 f(x\_2)$$

implying $f$ is convex.

Answer: 
Your reasoning is almost correct. However, you switch from $t\_i$ to $f(x\_i)$ without justification – you should start out with $f(x\_i)$ instead of $t\_i$ from the beginning; this is a special case of the general property you're using.
the idea you used shows $f$ is convex, for showing that $f$ is convex homogenous or equivalently $epi f$ is convex cone you need to show that

$$f(\theta\_1 x\_1 + \theta\_2 x\_2) \le \theta\_1 f(x\_1) + \theta\_2 f(x\_2)$$ for all $\theta\_{i} \ge 0 $