Let's see what the various integrals measure, starting with $\int_0^1 |f(x)| \, dx$. Imagine the function $f$ on the interval $[0,1]$, split into the parts where it's positive, and the parts where it's negative. The total (unsigned) area of all the parts is exactly $\int_0^1 |f(x)| \, dx$.
Next, \int_0^1 |f'(x)| \, dx. We know that \int_0^1 f'(x) \, dx = f(1) - f(0), since f' measures the amount by which $f$ is increasing. Similarly, |f'| measures the amount by which $f$ is changing. Divide $f$ into parts where it's increasing and parts where it's decreasing, and reflect each decreasing part starting at $x$ along $y = f(x)$. Now \int_0^1 |f'(x)| \, dx measure how far the modified function reaches. Suppose now that $f(0) = 0$. In that case, we have the easy bound |f| \leq \int_0^1 |f'(x)| \, dx on the interval $[0,1]$. Since $[0,1]$ has unit length, $\int_0^1 |f(x)| \, dx \leq \max_{x \in [0,1]} |f(x)|$, so in case $f(0) = 0$ we're done.
If $f(0) \neq 0$ then the bound \int_0^1 |f'(x)| \, dx can be terribly wrong. The extreme situation is when $f$ is constant, and then the latter integral is equal to zero. So we need to consider the final integral $\left| \int_0^1 f(x) \, dx \right|$. Recall we split $f$ into positive and negative parts. Add their signed areas and return the magnitude to get $\left| \int_0^1 f(x) \, dx \right|$. In general we may get large cancellation this way - indeed, a simple symmetric construction (like the sine function) has $\left| \int_0^1 f(x) \, dx \right| = 0$ while $\int_0^1 |f(x)| \, dx$ can be arbitrarily large. However, if $f$ never changes sign, then it's easy to see that the two integrals are equal.
So we divide into two cases. If $f$ never changes sign then $\int_0^1 |f(x)| \, dx = \left| \int_0^1 f(x) \, dx \right|$. Otherwise, assume without loss of generality that $f(0) > 0$. The problem with \int_0^1 |f'(x)| \, dx was that in its estimation of $\max_{x\in [0,1]} |f(x)|$ it was missing the contribution of $f(0)$. However, since $f$ crosses zero, there is somewhere a contribution of exactly $f(0)$, which does not contribute to $\max_{x\in [0,1]} |f(x)|$ (indeed, $f$ is rather diminishing its magnitude). So in that case it is again true that \max_{x\in [0,1]} |f(x)| \leq \int_0^1 |f'(x)| \, dx, and we're done.