The space of test functions is $𝒟(a, b)$ consists of all $\mathrm{C}^{∞}$ functions $ϕ:(a, b) → ℝ$ with compact support in $(a, b)$. A distribution on $(a, b)$ is a linear functional $u: 𝒟(a, b) → ℝ$ with the boundedness property: For each compact subset $K$ of $(a, b)$ there exist constants $c=c_{K} ≥ 0$, $m=m_{K} ∈ ℕ_{0}$ such that $|⟨u, ϕ⟩| ≤ c \sum_{k=0}^{m} \sup \left|ϕ^{(k)}\right|$ for all $ϕ ∈ 𝒟(a, b)$ with $\operatorname{supp}(ϕ) ⊆ K$. If we can choose the constant $m ∈ ℕ_{0}$ independently of $K$, then $u$ has order at most $m$. For $m=0, u$ has order 0 if $u$ has order at most 0, and for $m ∈ ℕ, u$ has order $m$ if it has order at most $m$, but not order at most $m-1$.
Let $K$ be a compact subset of $(a, b)$. Put $ξ=ρ_{ε} * 𝟏_{[a+ε, b-ε]}$ for $ε<\dist(K, ℝ ∖(a, b)) / 3$. Then $ξ ∈ 𝒟(a, b), ξ=1$ near $K$. Let $ϕ ∈ 𝒟(a, b)$ be supported in $K$ and note that $(\max |ϕ|-ϕ) ξ ∈ 𝒟(a, b)^{+}$, so by assumption $V((\max |ϕ|-ϕ) ξ) ≥ 0$. By linearity and since $ϕ ξ=ϕ$ we arrive at $V(ϕ) ≤V(ξ) \max |ϕ|$. Applying this to $-ϕ$ yields by linearity of $V$ the bound $V(ϕ) ≥-V(ξ) \max |ϕ|$, and consequently, $|V(ϕ)| ≤ c_{K} \max |ϕ|$ where $c_{K}=|V(ξ)|$. This is the boundedness property with $m=0$, and so $V$ is a distribution of order 0.
Let $u ∈ 𝒟'(a, b)$. Then $u'$ is defined for $ϕ ∈ 𝒟(a, b)$ by $⟨u', ϕ⟩=-⟨u, ϕ'⟩$. Since differentiation is a linear map of test functions, it follows by linearity of $-u$ that also $u'$ is linear. Using the boundedness property of $u$ we obtain the boundedness property of $u'$ : For a compact subset $K$ of $(a, b)$ let the constants $c=c_{K}, m=m_{K}$ be as above. Then if $ϕ ∈ 𝒟(a, b)$ is supported in $K$ also $ϕ'$ is supported there and so $$ \left|\left⟨u', ϕ\right⟩\right|=\left|\left⟨u, ϕ'\right⟩\right| ≤ c \sum_{k=0}^{m} \sup \left|ϕ^{(k+1)}\right| $$ This bound shows that $u' ∈ 𝒟'(a, b)$. It also shows that if $u$ has order $m ∈ℕ_{0}$, then $u'$ has order at most $m+1$. Next, if $u ∈ \mathrm{C}^{1}(a, b)$, then we consider $u$ as a distribution by the rule $⟨u, ϕ⟩=∫_{a}^{b} u ϕ \mathrm{d} x$. Now by integration by parts $⟨u,-ϕ'⟩=-∫_{a}^{b} u ϕ' \mathrm{d} x=-[u ϕ]_{x → a+}^{x → b-}+∫_{a}^{b} u' ϕ \mathrm{d} x=∫_{a}^{b} u' ϕ \mathrm{d} x$, consequently the distributional derivative of $u$ can be identified with the usual derivative $u'$ in this case.
The function $d$ is a 1-periodic function with $$ d(x)= \begin{cases}x &\text{if } x ∈\left[0, \frac{1}{2}\right) \\ 1-x &\text{if } x ∈\left(\frac{1}{2}, 1\right)\end{cases} $$ Note that it is piecewise $\mathrm{C}^{1}$, so we may perform partial integration as for $\mathrm{C}^{1}$ functions. Hereby we find $$ ⟨d', ϕ⟩=∫_{ℝ}(𝟏_{I}-𝟏_{J}) ϕ \mathrm{d} x $$ where the function $𝟏_{I}-𝟏_{J}$ is the a.e. defined usual derivative of $d$. Here $$ I=\bigcup_{j ∈ ℤ}\left(j, j+\frac{1}{2}\right) \text{ and } J=\frac{1}{2}+I $$
The function $d_{n}$ is piecewise $\mathrm{C}^{1}$ and $2^{-n}$-periodic. Because $0 ≤ d_{n}(x) ≤2^{-1-n}$ for all $x$, the Weierstrass $M$-test shows that the series defining $t$ is uniformly convergent. It is therefore the uniform limit of continuous functions and so is continuous. Because all $d_{n}$ in particular are 1-periodic, it follows that also $t$ is 1-periodic.
Since for $n ≥ m$ the function $d_{n}$ is $2^{-m}$-periodic and $v_{m}-u_{m}=2^{-m}$ we have $$ \frac{t(v_m)-t(u_m)}{v_{m}-u_{m}}=\sum_{n=0}^{m-1} \frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}} \tag{1} $$ For $n<m$ the function $d_{n}$ is affine with slope $± 1$ on the interval $[u_{m}, v_{m}]$, hence $\frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}}= ± 1$ as required. The sum in (1) can therefore not converge to a finite limit as $m →+∞$. Since $$ \frac{t(x_{0}+h)-t(x_{0})}{h}=\sum_{n=0}^{m-1} \frac{d_{n}(x_{0}+h)-d_{n}(x_{0})}{h}=\sum_{n=0}^{m-1} \frac{d_{n}(v_m)-d_{n}(u_m)}{v_{m}-u_{m}} $$ when $h=v_{m}-x_{0}$, it follows that $t$ isn't differentiable at $x_{0}$.
Since uniform convergence implies convergence in the sense of distributions the series defining $t$ in particular converges in $𝒟'(ℝ)$. Because differentiation in the sense of distributions is a continuous operation in $𝒟'(ℝ)$ we get $t'=\sum_{n=0}^{∞} d_{n}'$ in $𝒟'(ℝ)$. Here we get from (iii) above $d_{n}'=2^{-n}2^nd'(2^nx)=d'(2^nx)$, where the right-hand side is the a.e. defined usual derivative of $d$ evaluated at $2^{n} x$.

Fix $r>0$. Let $ϕ ∈ 𝒟(ℝ)$ with support in $[-r, r]$. Then we have for each $ε ∈(0, r)$ because $f(x) / x$ is a continuous and odd function on $ℝ ∖\{0\}$ that $$ \left(∫_{-r}^{-ε}+∫_{ε}^{r}\right) \frac{f(x)}{x} \mathrm{~d} x=0 $$ and consequently, \begin{aligned} \left|\left(∫_{-r}^{-ε}+∫_{ε}^{r}\right) \frac{f(x)}{x} ϕ(x) \mathrm{d} x\right| &=\left|\left(∫_{-r}^{-ε}+∫_{ε}^{r}\right) \frac{f(x)}{x}(ϕ(x)-ϕ(0)) \mathrm{d} x\right| \\ &≤\left(∫_{-r}^{-ε}+∫_{ε}^{r}\right)|f(x)| \frac{|ϕ(x)-ϕ(0)|}{|x|} \mathrm{d} x \\ &\stackrel{\text{FTC}}{≤}\left(∫_{-r}^{-ε}+∫_{ε}^{r}\right)|f(x)| \max \left|ϕ'\right| \mathrm{d} x \\ &≤ 2 ∫_{0}^{r}|f(x)| \mathrm{d} x ⋅\max \left|ϕ'\right|. \end{aligned} Read the inequality resulting from the second and fourth lines in the above display: take $ε ↘ 0$ and use Fatou's lemma to see that $f(x) \frac{ϕ(x)-ϕ(0)}{x}$ is integrable over $[-r, r]$. It therefore follows from Lebesgue's DCT that $$ \left⟨\pv\left(\frac{f(x)}{x}\right), ϕ\right⟩=∫_{-r}^{r} f(x) \frac{ϕ(x)-ϕ(0)}{x} \mathrm{~d} x $$ is well-defined and hereby also that $\pv\left(\frac{f(x)}{x}\right): 𝒟(ℝ) → ℝ$ is a well-defined linear functional. Since for a given compact set $K$ in $ℝ$ we can find $r>0$ such that $K ⊂[-r, r]$ it follows that the principal value integral satisfies the boundedness property: $$ \left|\left⟨\pv\left(\frac{f(x)}{x}\right), ϕ\right⟩\right| ≤ c \max \left|ϕ'\right|\text{ for all }ϕ ∈ 𝒟(K) $$ where $c=2 ∫_{0}^{r}|f(x)| \mathrm{d} x$. It is therefore a distribution on $ℝ$. Finally, when $ϕ(0)=0$ the function $f(x) ϕ(x) / x$ is integrable, so in this case the above yields $\left⟨\pv\left(\frac{f(x)}{x}\right), ϕ\right⟩=∫_{ℝ} f(x) ϕ(x) / x \mathrm{~d} x$ as required in that case.
The answer is no:
any continuous and even function $f: ℝ → ℝ$ with $f(0)=0$ for which $f(x) / x ∉ \mathrm{L}_\text{loc}^{1}(ℝ)$ is a counter-example
If $\pv\left(\frac{f(x)}{x}\right)$ were a regular distribution on $ℝ$, then we would have for some $g ∈ \mathrm{L}_\text{loc}^{1}(ℝ)$ that $\left⟨\pv\left(\frac{f(x)}{x}\right), ϕ\right⟩=∫_{ℝ} g ϕ \mathrm{d} x$ holds for all $ϕ ∈ 𝒟(ℝ)$. But then it follows in particular from (b)(i) that $∫_{ℝ ∖\{0\}} \frac{f(x)}{x} ϕ(x) \mathrm{d} x=∫_{ℝ ∖\{0\}} g(x) ϕ(x) \mathrm{d} x$ for all $ϕ ∈ 𝒟(ℝ ∖\{0\})$ and so by the fundamental lemma of the calculus of variations that $f(x) / x=g(x)$ a.e., a contradiction proving the claim.

For instance we can take $$ f(x)= \begin{cases}\frac{1}{|\log | x||+1} &\text{if } x ∈ ℝ ∖\{0\} \\ 0 &\text{if } x=0\end{cases} $$ Then $f: ℝ → ℝ$ is continuous, even and $f(0)=0$. But $$ ∫_{-1}^{1}\left|\frac{f(x)}{x}\right| \mathrm{d} x=2 ∫_{0}^{1} \frac{\mathrm{d} x}{(1-\log x) x}=2 ∫_{1}^{∞} \frac{\mathrm{d} y}{y}=+∞ $$ so $f(x) / x ∉ \mathrm{L}_\text{loc}^{1}(ℝ)$.