I agree with the claim that translation by an element of the quotient group, $T_{[g]}:G/H \longrightarrow G/H$ Is a continuous function from $G/H$ to $G/H$, when $H$ is a normal subgroup of $G$. However, for $G/H$ to be a topological group with that operation, one needs multiplication to be a continuous function from $G/H \times G/H$ to $G/H$.
Continuity of the multiplication map implies continuity of left and right multiplication maps, but I don't see why continuity of the left and right multiplication maps would imply the continuity of multiplication map $M : G/H \times G/H \longrightarrow G/H$
First, prove that the quotient map $q: G \to G/H$ via $q(x)=xH$ is an open map. I'm assuming you've already done this since you believe (rightly so) that $T_{[g]}$ is continuous. If not, take a moment to prove this via a homogeneity/translation argument. (By the way, $q$ is open even when $H$ isn't normal.)
Next, prove that $Q(x,y) = (xH, yH)$ defines an open continuous map $Q: G \times G \to G/H \times G/H$. This isn't hard, because $Q(U \times V) = q(U) \times q(V)$. Finally, write out the square diagram that says $$ q \circ \mu = M \circ Q $$ where $\mu: G \times G \to G$ is your original product. Check that this square commutes and use open-ness of $Q$ to prove that $M$ is continuous.
A similar trick shows that inversion on $G/H$ is continuous. Here, $q \circ i = I \circ q$ where $i$ is inversion on $G$ and $I$ is inversion on $G/H$.