How does this happen in RSA malleability? I don't understand how the $E(m)$ turns into $E(mt)$. I mean, I don't know how does that transformation happen and how does the equation occur. 

$$E(m) \cdot t^e \bmod n = (mt)^e \bmod n = E(mt)$$

See, $E(m)$ is $m^e \bmod n$, so if you plug in real values, it won't work.

For example:


$c_1 = 17^{23} \bmod 35 = 33$
$c_2 = 2^{23} \bmod 35 = 18$


Now, $c_1 \cdot c_2$ should result in $33 \cdot 18 = 594$, i.e. $E(mt)$.
But if you plug in the values in $(mt)^e \bmod n$, you get $34$ instead of $594$.

How are these equal?
As mikeazo notes in the comments, RSA operates on the ring $\mathbb Z / n\mathbb Z$ of integers modulo $n$, for a given modulus $n = pq$.  In this ring,

$$E(m) \cdot t^e \equiv m^e \cdot t^e \equiv (mt)^e \equiv E(mt)\ \pmod n.$$

In particular, for $n =  35$, $e = 23$,

$$17^{23} \cdot 2^{23} \equiv 33 \cdot 18 \equiv 594 \equiv 34 \equiv 34^{23}\ \pmod{35}.$$

The notation in your question,


  "$E(m) \cdot t^e \bmod n = (mt)^e \bmod n = E(mt)$",


apparently using $\bmod$ as a modulo operator, is somewhat confusing here, although it's technically correct if you interpret the left hand side as "$(E(m) \cdot t^e) \bmod n$".  I would much prefer the notation I've used above, which makes it clear that we're doing all arithmetic in the ring $\mathbb Z / n\mathbb Z$, and that we're thus free to reduce (or not to reduce) intermediate values modulo $n$ any time we want.