My syllabus says that $\hom_{\mathbb F_2}(\mathbb F_2^2,\mathbb F_2^2)$ contains 16 elements. I’m guessing this is because $\mathbb F_2^{2\times 2}$ contains 16 elements, but I don’t see why this should be the reason.
It happens to be that we also have three bases for $\mathbb F_2^2$, so shouldn’t there be $< 16$ linear maps? Some linear map could be represented by several matrix representations, i.e. some of the matrices in $\mathbb F_2^{2\times 2}$ are similar.
On
Let's consider $\mathrm{Hom}_F(V,W)$ where $F$ is a field, and $V$ and $W$ are vector spaces over $K$ with dimensions $m$ and $n$. Then $\mathrm{Hom}_F(V,W)$ is a vector space of dimension $mn$; it corresponds to the $m\times n$ matrices over $K$.
In this example $K=\Bbb F_2$ and $m=n=2$. So we get a vector space of dimension $4$ over $\Bbb F_2$. It has order $|\Bbb F_2|^4=2^4=16$.
The set $\hom_{\mathbb{F}_2}(\mathbb{F}_2^2, \mathbb{F}_2^2)$ can be identified with $M_{2}(\mathbb{F}_2)$ (matrices of size $2\times 2$ with entries from $\mathbb{F}_2$). The association is not hard as every linear transformation corresponds to a matrix after you fix a basis. If you believe this, then it can be seen that $|M_{2}(\mathbb{F}_2)|=2^4=16$.