From a category perspective, the natural definition for $T$ to be invertible here is that there exists $S\in L(W,W)$ such that $S\circ T=T\circ S=I$.
As usual, $S\circ T=I$ implies that $T$ is injective, and $T\circ S=I$ entails that $T$ is surjective. So $T$ invertible implies it is bijective.
Now assume that $T$ is bijective. Then we have $S=T^{-1}$ that fits the definition above, provided we can prove that it belongs to $L(W,W)$.
Let us write that $T$ is linear, namely
$$T(\lambda x+\mu y)=\lambda T(x)+\mu T(y)$$
for all $x,y\in W$, and $\lambda,\mu\in\mathbb{R}$.
This is true in particular for $x=T^{-1}(x')$ and $y=T^{-1}(y')$ for all $x',y'\in W$.
So
$$
T(\lambda T^{-1}(x')+\mu T^{-1}(y'))=\lambda TT^{-1}(x')+\mu TT^{-1}(y')=\lambda x'+\mu y'.
$$
It only remains to apply $T^{-1}$ to the LHS and the RHS of the above. This proves the linearity of $T^{-1}$.
So yes, bijectivity is equivalent to invertibility in $L(W,W)$.
I think we can prove a little stronger result that
$$ T \text{ is surjective } \implies T^{t} \text{ is injective}. $$
But let's do it by putting all the definitions first.
Let $ V $ and $ W $ be linear spaces over a field $ F $.
$T : V \mapsto W $ be an isomorphism i.e a linear transformation that is injective and surjective. Of course, as a bijective map, there is a map $ T^{-1} : W \mapsto V $ and one can prove that this map is itself a linear transformation.
Let $ l : V \mapsto F $ be a "linear functional" i.e. a linear transformation from $ V $ to the one-dimensional vector space $ F $.
The set $ V^{*}= \left \{ l| l : V \mapsto F \text{ is a linear functional } \right \} $ forms a vector space over $ F $, with addition and scalar multiplication defined in the obvious way. We call it the dual of $ V $. Similarly, let $ W^{*} $ denote the dual of $ W $.
If $ l \in W^{*} $, then the composition $ l \circ T : V \mapsto F $ is a linear functional over $ V$, and so we get an assignment, $ T^{t} $, which sends $ l \in W^{*} $ to $ m = l \circ T $ in $ V^{*} $. We can prove that $ T^{t} : W ^{ * } \mapsto V ^ { * } $ is a linear transformation, such that, for $ v \in V $, we have
$$ (T^{t}( l)) ( v ) = l ( T ( v ) ) = m(v). $$
We now prove the following:
(1) If $ T $ is surjective, then $ T^{t} $ is injective.
(2) If $ T $ is invertible, then $ T^{t} $ is surjective.
$ \textit{Proof:} $ We prove (1) first. Suppose that $ T^{t}(g_{1}) = T^{t} ( g_{2} ) $ where $ g_{1}, g_{2} \in W^{*}$. Then, $ g_{1}T = T^{t}(g_{1} ) = T^{t}(g_{2}) = g_{2}T $. Thus, for every $ v \in V $, we have that
$$ g_{1}(T(v)) = g_{2}(T(v)). $$
As $ T $ is surjective, every $ w \in W $ can be written as $ T(v) $ for some $ v \in V $, and so, the last equality says that $ g_{1} $ and $ g_{2} $ agree on every $ w \in W $, and so $ g_{1} = g_{2} $.
To prove (2), let $ m \in V^{*} $, then $ m \circ T^{-1} \in W ^ { * } $, and we have
$$ T^{t} ( m T^{ - 1} ) = (m T^{-1} ) T = m( T^{-1} T) = m ( I ) = m. $$
$ \textit{ Remark : } $ The only thing that I don't like about your proof is when you "apply" the linear functional $ ( g_{1} - g_{2})T $ to $ T^{t} $. You are composing two functions. Don't say apply. The functional $ ( g_{1} - g_{2} ) T $ can only be applied to vectors $ v \in V $. Otherwise, the proof seems fine.
Best Answer
If $A$ is invertible, then $m=n$.
Since you have proved that $T$ is injective, it follows from the rank-nullity theorem that $T$ is surjective because the two spaces have the same dimension (and $T$ is linear).