Is there a better notation for integrating twice?

No, there is no better notation - the double-integral notation is standard. However, the way you've written it is problematic. Notice that when you do an indefinite integral, you get a $+c$ at the end. This is a constant, so when integrated again we have $+ct$. Evaluating from $2$ to $5$, this gives a $+3c$ at the end of your answer - which you really don't want, since your answer should be a number.

In a double integral, the inner integral should a) always be definite and b) be with respect to a different variable than the outer integral. In your case, recall that velocity is not the indefinite integral of acceleration - it's $v_0 + \int_0^ta(s)ds$, where $t$ is the time. So what you want is $\int_2^5\int_0^ts^2dsdt$.

This distinction between $s$ and $t$ is important - without it, you'll run into ambiguities as to which $t$ each $dt$ applies to.

This does not quite make sense to me, due to $$ \int t^2 \,dt = \frac 13 t^3 + C $$ Your integral just gives $$ \int_2^5 \int_t \tau^2 \, d\tau\, dt = \frac 1{12} (5^4 - 2^4) + 3C $$ for some arbitrary $C$, hence the result is just any(!) number. The result $23$ is as fine as $\pi$, just different and equally justified choices of $C$.

So I'd stick to an double integral with limits, so variable limits on the inner one. It is misleading to use the same integration variable twice, the standard way around is to use a "look-alike" variable name, for example the corresponding greek letter. So, a possibility is to write $$ \int_2^5 \left(u + \int_0^t \tau^2 \,d\tau\right) \, dt $$ where $u$ is the initial velocity. And if you want to stick to indefinite limits, I'd write $$ \int_2^5 \int_t \tau^2 \,d\tau \, dt $$