Trouble understanding the question regarding dual spaces 
If $(\alpha_0,\alpha_1,\alpha_2,\dots)$ is an arbitrary sequence of
complex numbers, and if $x$ is an element of $\mathcal{P}$,
$x(t)=\sum\limits^n_{t=0}\xi_i t^i$, write
$y(x)=\sum^n_{i=0}\xi_i\alpha_i$. Prove that $y$ is an element of
$\mathcal{P'}$ and that every element of $\mathcal{P'}$ can be
obtained in this manner by a suitable choice of the $\alpha's$.

Here $\mathcal{P'}$ is a dual space of $\mathcal{P}$. Can you explain what the second question means?
 A: I’m not sure whether you only wanted clarification on the (second) question or a demonstration that it’s true. To answer them below, let’s write $\alpha=(\alpha_1,\alpha_2,\ldots)$ for the stated arbitrary infinite sequence. For any polynomial $x$, let $x^{(k)}$ be the $k$th derivative and write $Dx:=(x(0),x^{(1)}(0), \frac{x^{(2)}(0)}{2!},\ldots,\frac{x^{(n)}(0)}{n!},\ldots)$, which is simply the sequence coefficients of $x$ (consisting of an eventually trailing sequence of $0$s). Also, for a sequence $\delta=(\delta_1,\delta_2,\ldots)$, write $\delta\cdot\alpha:=\delta_1\alpha_1+\delta_2\alpha_2+\cdots$ for the dot product.
Now, observe that your space $\mathcal{P}$ is linearly isomorphic to the space $c_{00}$ of eventually zero sequences (consisting of all sequences which have only finitely many nonzero elements) by the canonical map
$$x\mapsto Dx\,.$$
It is not difficult to prove (I will supply details upon request, but you can mimic the situation for finite-dimensional spaces) that a linear functional $L$ in $c_{00}’$ is specifically of the form $$L(\delta)=\delta\cdot\alpha$$ for some infinite sequence $\alpha$ and all $\delta\in c_{00}$.
Thus via the isomorphism $\mathcal{P}\sim c_{00}$, then for the first question, the map
$$f\colon\mathcal{P}\to\mathbb{C}\,,~\,~\,~ x\mapsto Dx\cdot\alpha\,,$$
which gives you $fx=y$ in your notation, is a linear functional; and for the second question, every linear functional in $\mathcal{P}’$ is therefore of the form as $f$ above.
A: For $i\in \Bbb N_0$ and $t\in \Bbb C$ let $f_i(t)=t^i$ (with the usual convention that $0^0=1$).
The 2nd part  asks you to prove that if $y\in P'$ then there is a sequence $(\alpha_0,...)$ such that $y(x)=\sum_i\xi_i \alpha_i.$ Which should be obvious because  $\{f_i:i\in \Bbb N_0\}$ is a vector-space basis for $P.$ That is, $\alpha_i=y(f_i).$
