So the way I view canonical isomorphisms is an isomorphism which depends only the construction/definitions of the spaces we’re working with.
My question is then this: if I want to prove that there exists a canonical isomorphism between two vector spaces, I have to write down a map who doesn’t assume a basis, however, can I use a basis to show that this map is an isomorphism?
For example, if I want to find an isomorphism $V\otimes W^*\rightarrow Hom(W,V)$, then the mao is given on simple tensors by;
$$F:v\otimes \omega\longrightarrow T$$
Where $T$ is the linear transformation defined by:
$$T(u)=v\cdot\omega(u)$$
To show this is surjective and injective, can I use a basis for $V$ and $W^*$?
Yes you can. In general, a collection of maps $\alpha_A$ is natural in $A \in$ "some category", means that the maps $\alpha_A$ are defined uniformly across all objects $A$ varying in the category. In other words, you don't make any arbitrary choices while defining these maps, and all of them are defined in the same way. This is the concept of naturality/a natural transformation and is formalized by demanding the commutativity of some naturality squares.
Checking that $\alpha_A$ is an isomorphism for each $A$ is a completely independent issue.
Once you combine naturality of the maps $\alpha_A$ (in the variable $A$) and the fact that each component (that is, each $\alpha_A$) is an isomorphism, you get the concept of a natural isomorphism (that is, a canonical isomorphism).
In the present context, the map $v \otimes w \mapsto [u \mapsto w(u).v]$ from $V \otimes W^{*} \to Hom(W,V)$ is natural in both the variables $V$ and $W$ (meaning that if you take any maps $V \to V'$ and $W' \to W$, then the induced diagram
$\require{AMScd}$ \begin{CD} V \otimes W^{*} @>>> Hom(W,V)\\ @VVV @VVV\\ V' \otimes W'^{*} @>>> Hom(W',V') \end{CD}
will commute (suggesting that the horizontal arrows do not depend on arbitrary choices and are defined in the same way).
Now, checking that each horizontal arrow is bijective is a completely independent issue , and you can employ whatever tools you have at hand.
The end result is that the isomorphism $V\otimes W^{*} \xrightarrow{\sim} Hom(W,V)$ is canonical.