The first part of this exercise seems straightforward to me, but I'm a bit confused about the second part. In the case $\alpha$ is transcendental we know that $k(\alpha)$ is actually isomorphic to $k(t)$ so it must be referring to the case where $\alpha$ is algebraic. But in that case I'm not sure how to show that there might not be an onto homomorphism?
@Brian Shin has given the answer, but let me write out some more details.
So the basic reason is that a field homomorphism is either 0 or injective, by the consideration of its kernel as an ideal in the domain field.
In the case at hand, since $k(t)\to k(\alpha)$ can't be injective when $\alpha$ is algebraic by dimension reasons, the homomorphism must be 0.
Therefore, there is no such a thing as a "quotient field", although quotient groups and quotient rings are so important.
Incidentally, if $f\in k[x]$ is the minimal polynomial of $\alpha$, then $$ k[x]/(f)\overset\cong\to k(\alpha); \ [g(x)]\to g(\alpha). $$ So a quotient ring just happens to be a field, since $f$ is irreducible.
Also note that this natural homomorphism $k[x]\to k(\alpha)$ can't be extended to $k(x)$, because the natural candidate $g(x)/h(x)\to g(\alpha)/h(\alpha)$ is not defined when $h(\alpha)=0$, which happens in particular when $f|h$.