It can be often illuminating to forget some of the technology and go back to a simpler time. Axler's text is a favourite of mine. If you can do all the problems in that book, then you know your linear algebra, let me assure you.
Regarding $Hom$...
You have two vector spaces $U,V$ (lets say they're finite dimensional). In plain english, $Hom(U,V)$ is the set of all (linear) maps from $U$ to $V$. We're in a finite dimensional case so lets go ahead and pick bases for $U$ and $V$ so they each look like $k^n$ for (possibly different) $n$. Think of them as column vectors.
So, how do you map one set of column vectors to another: you use matrices! You know how to scale matrices, and add matrices. In other words, $Hom(U,V)$ has just been given the structure of a vector space all on its own. Its a big vector space consisting of matrices of the appropriate size (so multiplying them with column vectors from $U$ makes sense) and one basis for instance is given by putting $0$ everywhere in the matrix except in one spot.
You can write this down in fancier ways without choosing bases too: Given maps $\phi$ and $\psi$ from $U$ to $V$, we can (pointwise) add and scale them and we get new linear maps. It is a good exercise to figure out what the basis I described above in terms of matrices looks like without explicitly thinking about the matrices. It is going to be a distinguished linear transformation. Figure out what it is!
Another great exercise is to set $V = k$ and understand $U^*$, the dual vector space. What do the above matrices look like in this case? Can you see for yourself that $U\cong U^*$ once you picked a basis?
Tensor products are somewhat more subtle. I would suggest trying to understand them in terms of ``what they do''. The tensor product of $U$ and $V$ is built to be the most general machine that can handle linear maps from $V\times U$, that are in fact bilinear (linear in each variable). Sit on this idea for a while. Then look at how a tensor product is defined in terms of generators and relations.
Once you have, try and make sense of: $k^n\otimes k^m \cong k^{mn}$.
