So the hint of the exercise is that I should use Taylor's theorem (I'm not assuming $f$ is analytic).
Then it's clear that all odd derivatives of $f$ at $0$ is $0$ so in some sense $h$ can be defined by the Taylor coefficients $h_k = f_{2k}$.
More generally, can I show that if $f \in C^{2k}$ and is even, then there exists $h \in C^k$ with $h(t^2) = f(t)$?
Here's an answer I have came up with using the idea suggested by copper hat in terms of looking at decay of the remainder term. I am not sure if $h(t) =: f(\sqrt t)$ is actually $C^k$ if $f \in C^{2k}$ However, I claim that
Suppose $f$ is even and $f \in C^{3k+1}(\mathbb{R})$, then $h(t)=: f(\sqrt t) \in C^k([0,\infty))$
Proof: Apply Taylor's theorem to $f$ at $0$ to the $2k$ order (assuming $k \geq 1$.
Then $f(t) = \sum_0^{k} a_i t^{2i} + t^{2k+1}\int_0^1 f^{(2k+1)}(ts)(1-s)^{2k+1} ds$
Then for $t \geq 0$, $h(t) = f(\sqrt t) = \sum_0^k a_i t^i + t^{\frac{2k+1}{2}}\int_0^1 f^{(2k+1)}(\sqrt{t}s)(1-s)^{2k+1} ds$
Now for any $l \leq k$, if we take $l$th derivative then the remainder term will involve a positive power of $t$ times an integral expression with bounded integrand since we can still take up to $k$ derivative of $f^{(2k+1)}$. This can be seen using chain rule and product rule and differentiation under the integral. (Edit: To elaborate, when differentiating the remainder term, any derivative of the intgrand will decreases the total power of t of the remainder by $\frac{1}{2}$ by chain rule, while differentiating $t^{\frac{2k+1}{2}}$ will decrease that power by $1$, since I'm not differentiating more than $k$ times the total power of $t$ from differentiating the remainder term will be positive, hence tends to 0 as $t$ tends to 0)
So $\lim_{t \rightarrow 0^+} h^{(l)}(t)$ can be read off from the $l$th term in the polynomial part (the first sum)
QED
It is known that for every sequence there exists a smooth function whose Taylor series coefficients are equal to that sequence, consider the sequence with the first $k$ terms are $\lim_{t \rightarrow 0^+} \frac{1}{l!}h^{(l)}(t)$, then define $h(t)$ to be equal to the said function above for $t \leq 0$, then this $h$ would have the desired property $h(t^2) = f(t)$ and it is $C^k$.
Remark: I'm not sure if this is the sharpest result, but it is enough to show that if $f \in C^\infty$, then there exists such a $h \in C^\infty$.
Let us prove that if $k\ge1$ and $f:\Bbb R\to\Bbb R$ is even and $2k$ times differentiable (not necessarily $C^{2k}$), then $f(x)=h(x^2)$ for some $C^k$ function $h$ on $[0,+\infty)$ (which can then be extended to a $C^k$ function on $\Bbb R,$ thanks to Borel's lemma). The only problem is the regularity of $h$ at $0.$
For a fixed $k,$ consider the universal constants $a_0,\dots,a_k$ (we won't need to calculate them) such that for every $k$ times differentiable function $F:[0,+\infty)\to\Bbb R,$ the function $H:[0,+\infty)\to\Bbb R$ defined by $H(x^2)=F(x)$ satisfies: $$\forall x>0\quad H^{(k)}(x^2)=\frac1{x^{2k}}\sum_{i=0}^ka_ix^iF^{(i)}(x).$$ If $F$ is moreover $2k$ times differentiable, Taylor gives: $$\begin{align}H^{(k)}(x^2)&=\frac1{x^{2k}}\left(\sum_{i=0}^ka_ix^i\sum_{j=0}^{2k-i}\frac{F^{(i+j)}(0)}{j!}x^j+o(x^{2k})\right)\\ &=\frac1{x^{2k}}\left(\sum_{n=0}^{2k}c_nF^{(n)}(0)x^n+o(x^{2k})\right) \end{align}$$ where the (again universal!) constants $c_0,\dots,c_{2k}$ are given by: $$c_n=\sum_{i\le\min(k,n)}\frac{a_i}{(n-i)!}.$$ Now, since $f$ was assumed to be even, its derivatives of odd order vanish at $0,$ so that for our function $h(t):=f(\sqrt t),$ the previous identity simplifies to $$\forall x>0\quad h^{(k)}(x^2)=\frac1{x^{2k}}\left(\sum_{m=0}^kc_{2m}f^{(2m)}(0)x^{2m}+o(x^{2k})\right)$$ Assuming (by induction) that we already know that $h$ is $C^{k-1},$ we are done if we prove that $$\forall m<k\quad c_{2m}=0,$$ because this will entail the existence of $\lim_{0^+}h^{(k)}$ (equal to $c_{2k}f^{(2k)}(0)$), hence the continuous differentiability at $0^+$ of $h^{(k-1)}.$
Here comes the magic of the universality: to prove that $c_{2m}=0,$ simply apply the formula above to some $C^\infty$ function $h$ for which the $2m$-th differential at $0$ of $x\mapsto h(x^2)$ is non-zero (I leave it to you to produce such a function $h$), and use that $\lim_{x\to0}h^{(k)}(x^2)$ exists.
It is clear that for $t \ge 0$ that we must have $h(t) = f( \sqrt{t})$. Since $f$ is smooth and $t \mapsto \sqrt{t}$ is smooth for $t>0$, we see that $h$ is smooth on $t >0$.
Since $f$ is even, for any $n$, Taylor's theorem gives $f(t) = \sum_{k=0}^n {f^{(2k)}(0) \over (2k)!} t^{2k} +R_{2n+1}(t)$, where $R_k(t) = {t^{k+1} \over k!} \int_0^1 f^{(k+1)}(st) (1-s)^k ds$. Hence, for $t>0$ we have $h(t) = \sum_{k=0}^n {f^{(2k)}(0) \over (2k)!} t^{k} +R_{2n+1}(\sqrt{t})$. (Note that the subscript on $R$ is $2n+1$.)
(Thanks to @AnneBauval's meticulous attention for catching an overreaching conclusion at this point in an earlier version of the proof.)
Let $r(t) = R_{2n+1}(\sqrt{t})$, then $r(t) = {t^{n+1} \over (2n+1)!} \int_0^1 f^{(2n+2)}(s\sqrt{t}) (1-s)^{2n+1} ds$.
If we show that the one sided derivatives $r^{(k)}(0)$ exist and are zero for $k=0,...,n-1$ then it follows that the same is true for $h^{(k)}(0)$. Since $n$ is arbitrary, if follows that $h^{(k)}(0)$ exists for all $k$.
Let ${\cal T}_p$ be the span of the functions of the form $t \mapsto t^l \int_0^1 f^{(k+1)}(s\sqrt{t}) (1-s)^{k} s^m ds$, for $k \ge 0, l \ge p, m \ge 0$ (and $t \ge 0$, of course). The elements of ${\cal T}_p$ are continuous for $p \ge 0$.
If $p>0$ and $\phi \in {\cal T}_p$, note that $\phi(0) = 0$. If $p>1$ and $\phi \in {\cal T}_p$, note that $\phi$ is differentiable (a one sided derivative from the right at $t=0$), $\phi'(0) = 0$ and $\phi' \in {\cal T}_{p-1}$.
Since $r \in {\cal T}_{n+1}$, it follows that $r(0)=r^{(1)}(0) = \cdots = r^{(n-1)}(0) = 0$ and so $h$ has one sided derivatives at $t=0$ and these are given by $h^{(k)}(0) = k!{f^{(2k)}(0) \over (2k)!}$.
All that remains is to show that $h$ can be extended to a smooth function for $t <0$. The tool for this is Borel's Lemma applied to the sequence $h^{(k)}(0)$. This gives the existence of a smooth function $g$ such that $g^{(k)}(0)= h^{(k)}(0)$.
To finish, let $h(t) = \begin{cases} f(\sqrt{t}, & t>0,\\ g(t),& \text{otherwise}\end{cases}$
