If $f: A\rightarrow B$ and $g: B\rightarrow C$ are surjective, then $g\circ f$ is surjective.

This is bordering on too pedantic a marking. You could have said "by the definition of surjectivity, since $g:B\to C$ is surjective, for the given element $m\in C$ there exists an element $b\in B$ with $g(b)=m$". Similarly for the second "why?". Again, your answer is in fact very good, even perfect I would say. It clearly shows you understand the concepts and you solved it correctly. The marking is too pedantic, unless the instructions were to clearly state where and how you used each and every hypothesis.

The grader is being unfair unless this is supposed to be a pedantic proof class, but by definition is a rather poor turn of phrase to use.

You want by the surjectivity of $g$ and by the surjectivity of $f$.

An alternative would be by the surjectivity of $g$ and the definition of surjectivity and (the same for $f$).

Also, to be really clear and precise, you should break up your statement into justifying the existence of $b$ and then justifying the existence of $a$.

To wit:

"... there exists a $b \in B$ such that $g(b) = m$ by the surjectivity of $g$ ..."

and then the rest, using the surjectivity of $f$.

To be extremely precise, state the definition of surjectivity and use modus ponens:

Statement 1: $g$ is surjective

Statement 2: $g: B \rightarrow C$ is surjective implies for all $m \in C$ there exists a $b \in B$ such that $g(b) = m$.

Hence, by modus ponens, for all $m \in C$ there exists a $b \in B$ such that $g(b) = m$.

And so on.

Your answer is very good but I think whoever marked it meant that you need to say explicit where you use that $f$ and $g$ are surjective. For instance:

Assume $f$ and $g$ are surjective. Let $m$ be an element of $C$. Because $g$ is surjective, there exists a $b$ that's an element of $B$, such that $g(b) = m$ and, because $f$ is surjective, an $a$ element of $A$ such that $f(a) = b$ by definition. Thus, $(g \circ f)(a)= g(f(a)) = g(b)=m$, and so $g \circ f$ is surjective.