To directly show that $V$ maps bounded sets to precompact sets, the only idea I have would be to replicate most of the proof of Ascoli's theorem.
We can show it semi-directly, though, by exposing $V$ as the norm-limit of operators with finite-dimensional range.
For $1 \leqslant k < n$, let
$$\lambda_{n,k}(f) = \int_0^{k/n} f(t)\,dt,$$
and
$$\chi_{n,k}(x) = \begin{cases} 1 &, \frac{k}{n} \leqslant x < \frac{k+1}{n}\\ 0 &, \text{ otherwise}. \end{cases}$$
Define
$$V_n(f) = \sum_{k=1}^{n-1} \lambda_{n,k}(f)\cdot \chi_{n,k}$$
for $n \geqslant 1$. Then $V_n$ is a continuous operator with finite-dimensional range. For $0 \leqslant k < n$ and $\frac{k}{n} \leqslant x < \frac{k+1}{n}$, we have
$$\lvert V(f)(x) - V_n(f)(x)\rvert = \left\lvert\int_{k/n}^{x} f(t)\,dt \right\rvert
\leqslant \sqrt{x-\frac{k}{n}}\cdot \lVert f\rVert_{L^2} \leqslant \frac{1}{\sqrt{n}}\lVert f\rVert_{L^2},$$
and hence $\lVert V - V_n\rVert \leqslant \frac{1}{\sqrt{n}}$. Thus $V$ is the norm-limit of operators with finite-dimensional rank, therefore compact.
Lemma: Let $X,Y$ be Banach spaces. If $T_n:X\to Y$ are bounded operators with finite dimensional range and $T_n\to T\in B(X,Y)$ in operator norm, then $T$ is a compact operator.
For the case $X=Y=L^p$, $1< p\le\infty$, we will write $T$ as a norm limit of finite rank operators and deduce compactness of $T$ by the above lemma.
Let $n>1$ be an integer and partition the interval $[0,1]$ in $n$ subintervals of equal length, namely $I_{j,n}:=[\frac{j-1}{n},\frac{j}{n})$, $j=1,\dots,n$. Let $\phi_{j,n}$ denote the indicator function of $I_{j,n}$. Set
$$T_n:L^p[0,1]\to L^p[0,1], \;\;\;T_n(f)=\sum_{j=1}^n\bigg(\int_0^{\frac{j}{n}}f(t)dt\bigg)\cdot\phi_{j,n}$$
Note that the operator $T_n$ has automatically finite dimensional range, since $T_n(f)$ lies in the linear span of the functions $\{\phi_{j,n}:j=1,\dots,n\}\subset L^p[0,1]$. Also,
for $x\in[0,1]$ there exists a unique $j_x\in\{1,\dots,n\}$ such that $x\in I_{j_x,n}$ so $\phi_{i,n}(x)=1$ if and only if $i=j_x$. So
$$|T_n(f)(x)|^p=\bigg|\sum_{j=1}^n\bigg(\int_0^{\frac{j}{n}}f(t)dt\bigg)\cdot\phi_{j,n}(x)\bigg|^p=\bigg|\int_0^{\frac{j_x}{n}}f(t)dt\bigg|^p\le\bigg(\int_0^1|f(t)|dt\bigg)^p\le\|f\|_p^p$$
where in the last inequality we have used the fact that $\|f\|_1\le\|f\|_p$ for $f\in L^p[0,1]$. Thus
$$\|T_n(f)\|_p^p=\int_0^1|T_n(f)(x)|^pdx\le\int_0^1\|f\|_p^pdx=\|f\|_p^p$$
and thus $\|T_n\|\le1$, so $T_n$ are indeed bounded operators with finite rank.
We now show that $\|T_n-T\|_p\to0$. Indeed, we have
$$|T_n(f)(x)-T(f)(x)|^p=\bigg|\sum_{j=1}^n\int_0^{\frac{j}{n}}f(t)dt\cdot \phi_{j,n}(x)-\int_0^xf(t)dt\bigg|^p=$$ $$=\bigg|\int_0^\frac{j_x}{n}f(t)dt-\int_0^xf(t)dt\bigg|^p\;\;(\star)$$
where $j_x\in\{1,\dots,n\}$ is the unique integer such that $x\in I_{j_x,n}$ (and the above equality occurs because $\phi_{j_x,n}(x)=1$ and $\phi_{i,n}(x)=0$ for $i\ne j_x$). Continuing from $(\star)$, if we denote by $q$ the conjugate exponent $(1/p+1/q=1)$, we have
$$(\star)=\bigg|\int_x^{\frac{j_x}{n}}f(t)dt\bigg|^p\le\bigg|\int_0^1\chi_{I_{j_x,n}}(t)f(t)dt\bigg|^p\le$$ $$\le\bigg(\int_0^1\chi_{I_{j_x,n}}(t)\cdot|f(t)|dt\bigg)^p\le\bigg(\mu(I_{j_x,n})^{1/q}\cdot\|f\|_p\bigg)^p=\frac{1}{n^{p/q}}\cdot\|f\|_p^p$$
where we used Holder's inequality. Therefore,
$$\|T_n(f)-T(f)\|_p^p=\int_0^1|T_n(f)(x)-T(f)(x)|^pdx\le\int_0^1\frac{1}{n^{p/q}}\cdot\|f\|_p^pdt=\frac{1}{n^{p/q}}\cdot\|f\|_p^p$$
and thus $\|T_n-T\|_p\le\frac{1}{n^{1/q}}$. Letting $n\to\infty$ gives $T_n\to T$.
P.S:
Why it is reasonable to define the operators $T_n$ the way we did? First, we need them to have finite dimensional range. Second, we look at $T(f)(x)=\int_0^xf(t)dt$. This is a number very close to $\int_0^{j/n}f(t)dt$ for some suitable $j,n$. So it feels natural to partition the unit interval in small intervals of length $1/n$ and define $T_n(f)$ by the rule "take a $x$, determine in which small interval it lies (i.e. find the proper $j_x$), then assign the value $\int_0^{j_x/n}f(t)dx$. Implicitly we have been multiplying with $\phi_{j,n}$ and adding up, to make sure we evetually obtained the correct $j_x$. I hope this helps you understand the reasoning here.
Best Answer
As you said, we have to apply Ascoli-Arzelà theorem.
Of course the sequence $Tf_n$ is equally bounded, because $||f_n||\leq M$ for every $n$ and so
$$||Tf_n||\leq M sup_{x}(\int_0^xe^ydy)=M(e-1)$$
Moreover $Tf_n$ is uniformly equicontinuos because if you take an $\epsilon>0$ and $x_1\leq x_2\in [0,1]$ then
$|Tf_n(x_2)-Tf_n(x_1)|\leq M\int_{x_1}^{x_2}e^ydy=e^{x_2}-e^{x_1}\leq \epsilon$
for $|x_2-x_1|\leq \delta$ (here we are using that the function $e^x$ is uniformly continuos)