For a concrete counterexample, consider the following.
Define $f_n$ on $[0,1/n]$ by $ f_n (x) = \cos ^2 \bigg(\frac{\pi }{x}\bigg), \;\; x \in [\alpha_n,1/n], $ and $ f_n (x) = 0, \;\; x \in [0,\alpha_n], $ where $\alpha_n$ is very small compared to $1/n$, and $f_n(\alpha_n)=0$. Note that thus $0 \leq f_n \leq 1$, $f_n (0) = 0$, and $f_n (1/n) = 1$, as required. Then, restricted to the interval $[\alpha_n,1/n]$, f'_n (x) = 2\cos \bigg(\frac{\pi }{x}\bigg)\sin \bigg(\frac{\pi }{x}\bigg)\frac{\pi }{{x^2 }} = \sin \bigg(\frac{{2\pi }}{x}\bigg)\frac{\pi }{{x^2 }}, and, restricted to the interval $[0,\alpha_n]$, f'_n (x) = 0. Note that the issue of smoothness of $f_n$ and f'_n at the endpoints $\alpha_n$ and $1/n$ plays no role for our purposes. Now, define $\phi $ by $ \phi(x) = x\sin \bigg(\frac{{2\pi }}{x}\bigg) ,\;\; x \in (0,a], $ and $ \phi(0) = 0. $ Note that thus $\phi$ is continuous on $[0,a]$. Now, \int_0^{1/n} {f'_n (x)\phi (x)\,dx} = \int_{\alpha _n }^{1/n} {f'_n (x)\phi (x)\,dx} = \pi \int_{\alpha _n }^{1/n} {\frac{{\sin ^2 (\frac{{2\pi }}{x})}}{x} \,dx} . A change of variable $x \mapsto 1/x$ then gives \int_0^{1/n} {f'_n (x)\phi (x)\,dx} = \pi \int_n^{1/\alpha _n } {\frac{{\sin ^2 (2\pi x)}}{x}\,dx} . The right-hand side tends to $\infty$ if, for example, $1/\alpha_n > n^2$; then, in particular, \int_0^{1/n} {f'_n (x)\phi (x)\,dx} \not \to \phi (0).
EDIT: Particular choice of $\alpha_n$ is not essential here. This follows from the fact that, given any $M > 0$, $\int_M^x {\frac{{\sin ^2 (t)}}{t}\,dt} \to \infty$ as $x \to \infty$.
EDIT 2: In fact, $f_n \in C^1 [0,a]$. To show this, it suffices to show that \lim _{x \downarrow \alpha _n } f'_n (x) = 0 and \lim _{x \uparrow 1/n} f'_n (x) = 0 (this would imply that f'_n (\alpha_n) = 0 and $f'_n(1/n)=0$); indeed, \mathop {\lim }\limits_{x \downarrow \alpha _n } f'_n (x) = f'_n (\alpha _n + ) = 2\cos \bigg(\frac{\pi }{{\alpha _n }}\bigg)\sin \bigg(\frac{\pi }{{\alpha _n }}\bigg)\frac{\pi }{{\alpha _n^2 }} = 0 (since, by definition, $ \cos ^2 (\frac{\pi }{{\alpha _n }}) = f_n (\alpha_n) = 0$) and \mathop {\lim }\limits_{x \uparrow 1/n } f'_n (x) = f'_n \bigg(\frac{1}{n} - \bigg) = 2\cos \bigg(\frac{\pi }{{1/n }}\bigg)\sin \bigg(\frac{\pi }{{1/n }}\bigg)\frac{\pi }{{1/n^2 }} = 0 (since $\sin(n \pi)=0$).