The key point is that you're confusing uniform convergence and $L^2$ convergence ; indeed as $\mathcal{C}([0;1])$ is both a subspace of $\mathcal{B}([0;1])$ with $|.|_\infty$ and of $L^2([0;1])$ with $|.|_2$, you get two norms on the same vector space.
But as it isn't a finite-dimensional space, it can have non-equivalent norms - and indeed, those two norms definitely aren't equivalent, which in particular means that a sequence which has a good behaviour for the $L^2$ norm (the partial sums of the Fourier series) doesn't necessarily have a good $|.|_\infty$ behaviour.
EDIT: I should have said a little more ; there's an obvious inequality between the two norms (the mean inequality) so they are not that unrelated. But there is no reverse inequality, as can be shown by considering a sequence of piecewise linear functions : for $n\in\mathbb N$, consider $f_n$ as $t\mapsto n^\alpha-n^{\alpha+\beta}t$ on $[0;n^{-\beta}]$ and zero elsewhere ; if you choose $\alpha,\beta>0$ carefully, then you'll get a sequence which converges to zero for the $L^2$ norm, and won't converge uniformly.
The key point is that you're confusing uniform convergence and $L^2$ convergence ; indeed as $\mathcal{C}([0;1])$ is both a subspace of $\mathcal{B}([0;1])$ with $|.|_\infty$ and of $L^2([0;1])$ with $|.|_2$, you get two norms on the same vector space.
But as it isn't a finite-dimensional space, it can have non-equivalent norms - and indeed, those two norms definitely aren't equivalent, which in particular means that a sequence which has a good behaviour for the $L^2$ norm (the partial sums of the Fourier series) doesn't necessarily have a good $|.|_\infty$ behaviour.