Let $\Phi: \mathbb{P} _3 \to \mathbb{P} _1 \times \mathbb{R} \times \mathbb{R}$ be given by the assignment rule $\Phi(p)=(p'', p'(0), p(0))$.
I am asked:
Prove that $\Phi$ is a bijection.
Here's my attempt:
Let $p_1,p_2$ take on the general form $p_1=ax^3+bx^2+cx+d$ for some arbitrary $a,b,c,d\in \mathbb{R}$, and $p_2=ex^3+fx^2+gx+h$ for some arbitrary $e,f,g,h\in \mathbb{R}$. Assume that $f(p_1)=f(p_2)$. Therefore $p_1''=p_2''$, so by taking second derivatives, $6ax+2b=6ex+2f\implies a=e$ and $b=f$. Also, $p_1'(0)=p_2'(0)$, so $c=g$. Finally, $p_1(0)=p_2(0)$, so $d=h$. Since all of the coefficients of $p_1$ and $p_2$ are respectively equal, we can say that $p_1=p_2$, and thus $f(p_1)=f(p_2)\implies p_1=p_2$. Since these values for $p_1$ and $p_2$ were arbitrary, the function $\Phi$ is injective.
Let $(ax+b,c,d)\in \mathbb{P}_1 \times \mathbb{R} \times \mathbb{R}$ be generic. Choose $p=\frac{1}{6}ax^3+\frac{1}{2}bx^2+cx+d$. Then $$\Phi(p)=(\frac{d^2}{dx^2}(\frac{1}{6}ax^3+\frac{1}{2}bx^2+cx+d), \frac{1}{2}a\cdot0^2+b\cdot0+c, \frac{1}{6}a\cdot0^3+\frac{1}{2}b\cdot0^2+c\cdot0+d) = (ax+b,c,d)$$
Since $(ax+b,c,d)$ was generic, we can verify the statement $\forall(ax+b,c,d)\in \mathbb{P}_1 \times \mathbb{R} \times \mathbb{R},\exists p\in \mathbb{P}_3,\Phi(p)=(p'',p'(0),p(0))$. Therefore, the function $\Phi$ is surjective.
Since $\Phi$ is both injective and surjective, it must be bijective. $\blacksquare$
Did I miss anything? Thanks!
Your reasoning is perfectly fine and well done. There is just a couple of not serious remarks we can make on your wording.
We usually use the word arbitrary, which has the more accurate meaning of "choosen at your will" or "any element", instead of generic, that have a more "generic"(vague, wide) meaning of "not specified" or "undetermined". Also the arbitrariness is implicit in a sentence starting with "Let...". So you may reword to
(As a side note I would write "it follows that" instead of "we can verify the statement".)
Now about
I think you meant to write
because this is what you shown and $\Phi(p)=(p'', p'(0), p(0))$ holds for any $p$, since it is the very definition of $\Phi$.
And finnaly
It is not that it must be bijective. It is that it is bijective by definition.
I don't know if you are familiar with linear algebra but here is an alternative proof using linear algebra so you can study the text.
Proof. Since $\mathbb{P} _3$ and $\mathbb{P} _1 \times \mathbb{R} \times \mathbb{R}$ with the usual operations of sum and scalar multiplication are both $4$-dimensional vector spaces it is enough to show that $\Phi$ is linear and $\Phi(p) = \vec{0} \implies p = 0$ because for a linear transformation between vector spaces of same dimensions it holds that $$\Phi \text{ bijective} \iff \Phi \text{ injective} \iff \Phi^{-1}(\{0\}) = \{0\} \iff \left(\Phi(p) = \vec{0} \implies p = 0\right)$$
Thus, let $\lambda \in \mathbb{R}$ and $p,q \in \mathbb{P}_3$. So $$\begin{align} \Phi(\lambda p + q) &= ((\lambda p + q)'', (\lambda p + q)'(0), (\lambda p + q)(0))\\ &= (\lambda p'' + q'', \lambda p'(0) + q'(0), \lambda p(0) + q(0))\\ &= (\lambda p'', \lambda p'(0), \lambda p(0)) + (q'', q'(0), q(0))\\ &= \lambda(p'', p'(0), p(0)) + (q'', q'(0), q(0))\\ &= \lambda \Phi(p) + \Phi(q) \end{align}$$
Therefore $\Phi$ is linear. Now suppose that $\Phi(p) = \vec{0}$. Then $p'' = 0$, $p'(0) = 0$ and $p(0) = 0$, whence $\forall x \in \mathbb{R}$ it follows that $$ p'(x) = p'(0) + \int_0^x p''(x)\ dx = 0 $$ and $$ p(x) = p(0) + \int_0^x p'(x)\ dx = 0 $$
Hence $p = 0$. $\tag*{$\blacksquare$}$
Notice that the fact that $p$ is a polynomial is not used anywhere in this proof. This means that the bijectivity of $\Phi$ holds for more general spaces. For example,the space of the two times differentiable functions between $\mathbb{R}$ and itself: $\mathscr{D}^2(\mathbb{R}, \mathbb{R})$ rather than $\mathbb{P}_3$ and the space of the continuous functions $C(\mathbb{R}, \mathbb{R})$ rather than $\mathbb{P}_1$. (Actually it is possible to consider even more general spaces.)