Motivation behind proof
One idea could be to prove that the Taylor-like quadratic polynomial $$ g(s)=f(x)+f'(x)s+\frac{L}{2}s^2 $$ is an upper bound for $f(x+s)$. As $f$ is positive, so $g$ has to be positive, thus the minimum value of $g$, which by the vertex properties of quadratic functions is $$ g\left(-\frac{f'(x)}{L}\right)=\frac1{2L}\left(2Lf(x)-f'(x)^2\right), $$ has still to be positive.
By the mean value theorem, with some $θ\in(0,1)$, $$ f(x+s)-g(s)=(f'(x+\theta s)-g'(θs))s=(f′(x+θs)−f′(x)-Lθs)s. $$ As the difference of the derivatives is bounded by $Lθ|s|$, one gets $$ (f′(x+θs)−f′(x))s\le Lθs^2, $$ so that the whole expression is always non-positive, $$ f(x+s)\le g(s) $$
This is pretty much equivalent to the official solution, but from a somewhat different perspective. Maybe it will help you.
First, notice that both the given condition - let's label that as $(1)$ - and the one we want to prove - call that $(2)$ - are invariant under horizontal translations $x \rightarrow x+a$, so it suffices to prove $(f'(0))^2 < 2Lf(0)$ for any such function $f$. Then if we suppose that some $f$ satisfies $(1)$, but fails to meet $(2)$ somewhere else, say at $x = a$, just consider the function $g(x) = f(x+a)$; it also satisfies $(1)$ but fails to meet $(2)$ exactly at $x = 0$, which will be a contradiction once we prove this "restricted" version of $(2)$.
The approach to the proof is also by contradiction. Suppose that there exists some $f$ such that $(f'(0))^2 \geq 2Lf(0)$. Since $f(0) > 0$, this implies that either $f'(0) \geq \sqrt{2Lf(0)}$ or $f'(0) \leq -\sqrt{2Lf(0)}$.
Consider the first case: at $x = 0$, the function $f$ is increasing at a rate of $f'(0)$, and since $f'$ is continuous, the function must have been increasing over some interval to the left of $x = 0$. The idea is to show that the "net" increase over that interval would have to be greater than (or equal to) $f(0)$ which would be a contradiction because it would mean that $f$ must have been less than (or equal to) $0$ at the left endpoint.
So, a lower bound on the net increase, which is an integral of the derivative, will come from integrating a lower bound on the derivative. This we get from $(1)$ (applied to any non-positive $x$ and $y = 0$): $f'(x) \geq f'(0) + Lx$. We want this lower bound to be non-negative, so that we are certain $f$ is increasing, and that tells us to consider the interval $x \in \left[-{f'(0)\over L}, 0\right]$. Integrating yields $$ \int_{-f'(0)/L}^0 (f'(0)+Lx)dx = \frac{(f'(0))^2}{2L} \geq f(0) $$ by our assumption and the contradiction is achieved.
The case $f'(0) \leq -\sqrt{2Lf(0)}$ is analogous, just consider an interval to the right of $x = 0$ because now $f$ is decreasing at $x = 0$ and will be decreasing "for a while." We can show that it will have to "dip" at least as low as $0$ before the decrease can stop.