Split into the cases $0 < t \leq T$ and $t > T$, to conclude that
$$
(s*h)(t) = \int_0^{\min \{ t,T\} } {1 \cdot h(t - \tau )d\tau } = \int_0^{\min \{ t,T\} } {\alpha e^{ - \alpha (t - \tau )} d\tau } .
$$
EDIT: It follows that
$$
(s*h)(t) = 1 - e^{-\alpha t}, \;\; 0 < t \leq T,
$$
$$
(s*h)(t) = e^{ - \alpha (t - T)} - e^{ - \alpha t} ,\;\; t > T.
$$
(Since $(s*h)(t) = 0$ for $t \leq 0$, we see that $(s*h)(t)$ is continuous on $\mathbb{R}$.)
Relation to probability theory: With $s$ and $h$ as above, let $X$ be an exponential random variable with density function $h$, and $Y$ an independent uniform$[0,T]$ random variable, so that $Y$ has density function $\tilde s = s/T$.
By the law of total probability, conditioning on $Y$, we have
$$
{\rm P}(X + Y \le t) = \int_0^T {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau }.
$$
It follows that if $0 < t \leq T$, then
$$
{\rm P}(X + Y \le t) = \int_0^t {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau } = \frac{1}{T}\int_0^t {(1 - e^{ - \alpha (t - \tau )} )d\tau } = \frac{{t - (1 - e^{ - \alpha t} )/\alpha }}{T},
$$
while if $t > T$, then
$$
{\rm P}(X + Y \le t) = \int_0^T {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau } = \frac{1}{T}\int_0^T {(1 - e^{ - \alpha (t - \tau )} )d\tau } = \frac{{T - (e^{ - \alpha (t - T)} - e^{ - \alpha t} )/\alpha }}{T}.
$$
Hence, the density function of $X+Y$ is given by
$$
f_{X+Y} (t) = \frac{{1 - e^{ - \alpha t} }}{T} ,\;\; 0 < t \leq T,
$$
$$
f_{X+Y} (t) = \frac{{e^{ - \alpha (t - T)} - e^{ - \alpha t} }}{T}, \;\; t > T.
$$
On the other hand, since $X$ and $Y$ are independent with respective densities $h$ and $\tilde s \,(=s/T)$,
$$
f_{X+Y} (t) = (h*\tilde s)(t) = (\tilde s * h)(t) = \frac{{(s*h)(t)}}{T},
$$
from which it follows that
$$
(s*h)(t) = 1 - e^{ - \alpha t} ,\;\; 0 < t \leq T,
$$
$$
(s*h)(t) = e^{ - \alpha (t - T)} - e^{ - \alpha t} ,\;\; t > T
$$
(as we have already seen above).
When faced with a problem in Calculus involving piecewise functions (such as the Heaviside), you can almost always make it easier on yourself but dividing into cases where the function takes different simpler non-piecewise forms. Then evaluate those individually.
Another way to approach it in this case: we are guaranteed that $0 < \tau < t$ at all times, just due to the integration bounds. This means that $t - \tau > 0$ for the whole integration, so $H(t - \tau) = 1$. This reduces it to
$$\int_0^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau$$
At this point, as you stated -- if $\tau < a$, then the integrand vanishes. And for $\tau > \alpha$, the $H(\tau - \alpha)$ becomes 1. So we can break it up into these two parts. (As I said, case-splitting!)
$$ \int_0^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau = $$
$$\int_0^\alpha \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau + \int_\alpha^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} H(\tau-\alpha) d\tau =$$
$$\int_\alpha^t \frac{\tau}{\sqrt{\tau^2-\alpha^2}} d\tau$$
which should work out fine. (Keep in mind that this final form is only true if $\alpha>0$, though! )
Best Answer
In the following derivation, we will denote the various Fourier Transforms as follows $$F(s) = \mathcal{F}\left\{f(t)\right\}$$ $$G(s) = \mathcal{F}\left\{g(t)\right\}$$ $$H(s) = \mathcal{F}\left\{h(t)\right\}$$ $$K(s) = \mathcal{F}\left\{k(t)\right\}$$
Assuming all of the integrals below and Fourier Transforms exist, and generalizing $2t-\tau$ to $at-\tau$,
$$\begin{align*}F(s) &= \mathcal{F}\left\{f(t)\right\}\\ \\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(\tau)h(at-\tau)k(t-\tau)\mathrm{d}\tau\space e^{-2\pi i ts} \mathrm{d}t\\ \\ &=\int_{-\infty}^{\infty}g(\tau)\int_{-\infty}^{\infty}h(at-\tau)k(t-\tau)\space e^{-2\pi i ts} \mathrm{d}t\space \mathrm{d}\tau\\ \\ &=\int_{-\infty}^{\infty}g(\tau)\space\mathcal{F}\left\{h(at-\tau)k(t-\tau)\right\}\space \mathrm{d}\tau\\ \\ &=\int_{-\infty}^{\infty}g(\tau)\space\left[\mathcal{F}\left\{h(at-\tau)\right\}*\mathcal{F}\left\{k(t-\tau)\right\}\right]\space \mathrm{d}\tau\\ \\ &=\int_{-\infty}^{\infty}g(\tau)\space\left[e^{-2\pi i \tau\frac{s}{a}}\dfrac{1}{|a|}H\left(\dfrac{s}{a}\right)*e^{-2\pi i \tau s}K(s)\right]\space \mathrm{d}\tau\\ \\ &=\int_{-\infty}^{\infty}g(\tau)\int_{-\infty}^{\infty}e^{-2\pi i \tau\frac{\nu}{a}}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)e^{-2\pi i \tau (s-\nu)}K(s-\nu)\space \mathrm{d}\nu\space \mathrm{d}\tau\\ \\ &=\int_{-\infty}^{\infty}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)K(s-\nu)\int_{-\infty}^{\infty}g(\tau)e^{-2\pi i \tau\frac{\nu}{a}}e^{-2\pi i \tau (s-\nu)}\space \mathrm{d}\tau\space \mathrm{d}\nu\\ \\ &=\int_{-\infty}^{\infty}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)K(s-\nu)\int_{-\infty}^{\infty}g(\tau)e^{-2\pi i \tau\left(s-\frac{a-1}{a}\nu\right)}\space \mathrm{d}\tau\space \mathrm{d}\nu\\ \\ &=\int_{-\infty}^{\infty}\dfrac{1}{|a|}H\left(\dfrac{\nu}{a}\right)K(s-\nu)G\left(s-\frac{a-1}{a}\nu\right)\space \mathrm{d}\nu\\ \\ \end{align*}$$
So for $a = 2$
$$ F(s) =\int_{-\infty}^{\infty}\dfrac{1}{2}H\left(\dfrac{\nu}{2}\right)K(s-\nu)G\left(s-\frac{\nu}{2}\right)\space \mathrm{d}\nu$$
You are still left with a "convolution-like" integral, so there is not much to be gained in the general case. Although if any of $G(s), H(s), K(s)$ have a special form, such as a dirac delta function or a rectangle function, there may be some benefit.