In general, such an operator need not be compact. Indeed, let $g \in L^1 ((0,\infty)) $ be arbitrary and define
$$
\alpha (t,s) = g (t) \sum_{n=1 }^\infty 1_{[n,n+1)}(s) e^{2\pi i n t} $$
and consider the sequence $f_n = 1_{[n,n+1)} $ which is bounded in $L^1$. Then $B f_n (t) = g (t) e^{2\pi i n t} $, which converges weakly in $L^1$ to $0$, but has norm $\|B f_n \|_1 = \| g\|_1$ for all $n $.
Hence, $(B f_n) _n $ does not have a convergent subsequence, so that $B $ is not compact.
As an extension of the answer above, here is a complete characterization of those functions $\alpha$ which yield a compact operator $B$.
In the following, I will assume that $\alpha$ is measurable with
respect to the Borel-$\sigma$-algebra on $\left(0,\infty\right)\times\left(0,\infty\right)$.
I claim that the operator $B$ is compact if and only if there is
a null-set $N\subset\left(0,\infty\right)$ such that the set
$$
K_{0}:=\left\{ \alpha\left(\cdot,s\right)\,:\, s\in\left(0,\infty\right)\setminus N\right\} \subset L^{1}\left(\left(0,\infty\right)\right)
$$
is totally bounded, i.e., $K:=\overline{K_{0}}\subset L^{1}\left(\left(0,\infty\right)\right)$
is compact. Once this characterization is known, you can use the result
linked in the answer by @DavideGiraudo to obtain a further simplification.
Indeed, the assumption $|\alpha (t,s)| \leq \beta (t) $ ensures that $K_0$ always satisfies the 1-tightness condition (condition (ii) of the paper by Hanche-Olsen and Holden) and is bounded (condition (i)). Finally, if we extend each $\alpha (\cdot , s) $ to all of $\Bbb {R} $ by seting it to zero outside $(0,\infty) $, then condition (iii) becomes
$$
\inf_{\delta >0} \sup_{|y| < \delta} \mathrm {esssup}_{s \in (0,\infty)} \int_{\Bbb {R}} |\alpha (t+y, s) - \alpha (t,s)| dt = 0. \qquad (\dagger)
$$
The condition given by @DavideGiraudo implies this (more general) condition as follows:
\begin{eqnarray*}
& & \int_{\mathbb{R}}\left|\alpha\left(t+y,s\right)-\alpha\left(t,s\right)\right|dt\\
& = & \int_{-\delta}^{0}\left|\alpha\left(t+y,s\right)\right|dt+\int_{0}^{R}\left|\alpha\left(t+y,s\right)-\alpha\left(t,s\right)\right|dt+\int_{R}^{\infty}\left|\alpha\left(t+y,s\right)-\alpha\left(t,s\right)\right|dt\\
& \leq & \int_{-\delta}^{0}\beta\left(t+y\right)dt+R\cdot\sup_{\left|t-t'\right|<\delta}\sup_{s\geq0}\left|\alpha\left(t,s\right)-\alpha\left(t',s\right)\right|+2\int_{R}^{\infty}\beta\left(t\right)dt\\
\left({\scriptstyle \text{for }R\text{ large enough}}\right) & < & \int_{y-\delta}^{y}\beta\left(t\right)dt+R\cdot\sup_{\left|t-t'\right|<\delta}\sup_{s\geq0}\left|\alpha\left(t,s\right)-\alpha\left(t',s\right)\right|+2\varepsilon\\
\left({\scriptstyle \text{for }\delta=\delta\left(R\right)\text{ small enough}}\right) & \leq & \int_{-2\delta}^{\delta}\beta\left(t\right)dt+3\varepsilon\\
\left({\scriptstyle \text{for }\delta\text{ even smaller}}\right) & \leq & 4\varepsilon.
\end{eqnarray*}
Using a similar (but easier) argument, one can also show that it suffices to restrict the integral in $(\dagger)$ to $\int_0^\infty$ instead of $\int_{\Bbb{R}}$.
It remains to prove the claimed equivalence from above. Let us show both implications separately.
"$\Rightarrow$": Here, we assume that $B$ is compact. Hence,
the set $L:=\overline{B\left(B_{1}\left(0\right)\right)}\subset L^{1}\left(\left(0,\infty\right)\right)$
is compact, where $B_{1}\left(0\right)\subset L^{1}\left(\left(0,\infty\right)\right)$
denotes the closed unit ball. It is not hard to see that the map
$$
\alpha:\left(0,\infty\right)\to L^{1}\left(\left(0,\infty\right)\right),s\mapsto\alpha\left(\cdot,s\right)
$$
is measurable. Indeed, $L^{1}\left(\left(0,\infty\right)\right)$
is separable and for arbitrary $g\in L^{\infty}\left(\left(0,\infty\right)\right)\cong\left[L^{1}\left(\left(0,\infty\right)\right)\right]^{\ast}$,
the map
$$
s\mapsto\left\langle g,\alpha\left(\cdot,s\right)\right\rangle =\int_{0}^{\infty}\alpha\left(t,s\right)\cdot g\left(t\right)dt
$$
is measurable as a consequence of the Fubini theorem, since $\left|\alpha\left(t,s\right)\cdot g\left(t\right)\right|\leq\left\Vert g\right\Vert _{L^{\infty}}\cdot\beta\left(t\right)\in L^{1}\left(\left(0,\infty\right)\right)$.
Now, measurability of the map $\alpha$ defined above follows from
Pettis's theorem (cf. https://en.wikipedia.org/wiki/Weakly_measurable_function).
Using this measurability, it is not hard to see that we have
$$
Bf=\int_{0}^{\infty}\alpha\left(\cdot,s\right)\cdot f\left(s\right)ds\qquad\forall f\in L^{1}\left(\left(0,\infty\right)\right),
$$
where the right-hand side is a Bochner integral. Note that $\alpha\in L^{\infty}\left(\left(0,\infty\right);L^{1}\left(\left(0,\infty\right)\right)\right)\hookrightarrow L_{{\rm loc}}^{1}\left(\left(0,\infty\right);L^{1}\left(\left(0,\infty\right)\right)\right)$,
so that the Lebesgue differentiation theorem yields a null-set $N\subset\left(0,\infty\right)$
such that for all $s_{0}\in\left(0,\infty\right)\setminus N$, we
have
$$
\alpha\left(\cdot,s_{0}\right)=\lim_{\theta\downarrow0}\frac{1}{2\theta}\cdot\int_{s_{0}-\theta}^{s_{0}+\theta}\alpha\left(\cdot,s\right)ds=\lim_{\theta\downarrow0}B\left[\frac{1}{2\theta}\cdot\chi_{\left(s_{0}-\theta,s_{0}+\theta\right)}\right]\in\overline{B\left(B_{1}\left(0\right)\right)}=L.
$$
All in all, we have shown that the set $K_{0}$ defined above satisfies
$K_{0}\subset L$ and is hence totally bounded.
To see that the Lebesgue differentiation theorem also applies to vector-valued
functions, assume that $\varrho\in L_{{\rm loc}}^{1}\left(\mathbb{R}^{d};X\right)$,
where $X$ is a separable Banach space. Let $\left\{ x_{n}\,:\, n\in\mathbb{N}\right\} \subset X$
be dense. For each $n\in\mathbb{N}$, the function $t\mapsto\left\Vert \varrho\left(t\right)-x_{n}\right\Vert $
is in $L_{{\rm loc}}^{1}\left(\mathbb{R}^{d};\mathbb{R}\right)$,
so that the usual form of Lebesgue's differentiation theorem yields
a null-set $N_{n}\subset\mathbb{R}^{d}$ such that $\left\Vert \varrho\left(t\right)-x_{n}\right\Vert =\lim_{r\downarrow0}\frac{1}{\lambda\left(B_{r}\left(t\right)\right)}\int_{B_{r}\left(t\right)}\left\Vert \varrho\left(s\right)-x_{n}\right\Vert ds$
for all $t\in\mathbb{R}^{d}\setminus N_{n}$. Now, let $N:=\bigcup_{n\in\mathbb{N}}N_{n}$
and let $t\in\mathbb{R}^{d}\setminus N$ be arbitrary and $\varepsilon>0$.
There is some $n\in\mathbb{N}$ with $\left\Vert \varrho\left(t\right)-x_{n}\right\Vert <\varepsilon$.
Hence,
\begin{eqnarray*}
& & \frac{1}{\lambda\left(B_{r}\left(t\right)\right)}\int_{B_{r}\left(t\right)}\left\Vert \varrho\left(t\right)-\varrho\left(s\right)\right\Vert ds\\
& \leq & \frac{1}{\lambda\left(B_{r}\left(t\right)\right)}\left[\int_{B_{r}\left(t\right)}\left\Vert \varrho\left(t\right)-x_{n}\right\Vert ds+\int_{B_{r}\left(t\right)}\left\Vert x_{n}-\varrho\left(s\right)\right\Vert ds\right]\\
& \leq & \varepsilon+\frac{1}{\lambda\left(B_{r}\left(t\right)\right)}\int_{B_{r}\left(t\right)}\left\Vert x_{n}-\varrho\left(s\right)\right\Vert ds\xrightarrow[r\downarrow0]{}\varepsilon+\left\Vert \varrho\left(t\right)-x_{n}\right\Vert <2\varepsilon,
\end{eqnarray*}
which shows $\frac{1}{\lambda\left(B_{r}\left(t\right)\right)}\int_{B_{r}\left(t\right)}\left\Vert \varrho\left(t\right)-\varrho\left(s\right)\right\Vert ds\xrightarrow[r\downarrow0]{}0$
and in particular $\varrho\left(t\right)=\lim_{r\downarrow0}\frac{1}{\lambda\left(B_{r}\left(t\right)\right)}\int_{B_{r}\left(t\right)}\varrho\left(s\right)ds$.
"$\Leftarrow$": Here, we assume that there is a null-set $N\subset\left(0,\infty\right)$
such that the set $K_{0}$ from above is totally bounded, so that
$K=\overline{K_{0}}$ is compact. Hence, so is $K_{1}:=\left\{ \theta\cdot f\,:\,\left|\theta\right|\leq1\text{ and }f\in K\right\} $,
since $\mathbb{K}\times L^{1}\left(\left(0,\infty\right)\right)\to L^{1}\left(\left(0,\infty\right)\right),\left(\theta,f\right)\mapsto\theta\cdot f$
is continuous, where $\mathbb{K}\in\left\{ \mathbb{R},\mathbb{C}\right\} $.
Hence, also the closed convex hull $K_{2}:=\overline{{\rm co}\, K_{1}}\subset L^{1}\left(\left(0,\infty\right)\right)$
is compact, see e.g. Theorem 5.35 in the book "Infinite Dimensional
Analysis" by Aliprantis and Border.
I claim that $B$ maps the closed unit ball $B_{1}\left(0\right)\subset L^{1}\left(\left(0,\infty\right)\right)$
into $K_{2}$. This is clearly sufficient for compactness of $B$.
Assume that this fails. Then there is some $f\in B_{1}\left(0\right)$
such that $g:=Bf\notin K_{2}$. Since $K_{2}$ is compact and convex,
as is $\left\{ g\right\} $, the "Strong Separation Hyperplane Theorem"
(Theorem 5.79 from the same book) yields a nonzero, continuous linear
function $\varphi\in\left[L^{1}\left(\left(0,\infty\right)\right)\right]^{\ast}$
which strongly separates $K_{2}$ and $\left\{ f\right\} $. $\varphi$
is given by integration against some function $h\in L^{\infty}\left(\left(0,\infty\right)\right)$.
Strong separation means that there is some $\theta\in\mathbb{R}$
and $\varepsilon>0$ satisfying $\varphi\left(g\right)\geq\theta+\varepsilon$
and $\varphi\left(F\right)\leq\theta$ for all $F\in K_{2}$. Note
that $0\in K_{1}\subset K_{2}$, so that $0=\varphi\left(0\right)\leq\theta$.
Now note
\begin{eqnarray*}
\theta+\varepsilon & \leq & \varphi\left(g\right)=\varphi\left(Bf\right)\\
& = & \int_{0}^{\infty}h\left(t\right)\cdot\int_{0}^{\infty}\alpha\left(t,s\right)\cdot f\left(s\right)ds\, dt\\
\left(\text{Fubini}\right) & = & \int_{0}^{\infty}f\left(s\right)\cdot\int_{0}^{\infty}\alpha\left(t,s\right)\cdot h\left(t\right)dt\, ds\\
& = & \int_{\left\{ s\,:\, f\left(s\right)\neq0\right\} }\left|f\left(s\right)\right|\cdot\int_{0}^{\infty}\alpha\left(t,s\right)\cdot\frac{f\left(s\right)}{\left|f\left(s\right)\right|}\cdot h\left(t\right)dt\, ds\\
& = & \int_{\left\{ s\,:\, f\left(s\right)\neq0\right\} }\left|f\left(s\right)\right|\varphi\left(\underbrace{\alpha\left(\cdot,s\right)\cdot\frac{f\left(s\right)}{\left|f\left(s\right)\right|}}_{\in K_{1}\subset K_{2}\text{ for }s\in\left(0,\infty\right)\setminus N}\right)\, ds\\
& \leq & \int_{\left\{ s\,:\, f\left(s\right)\neq0\right\} }\left|f\left(s\right)\right|\cdot\theta\, ds\\
& = & \theta\cdot\left\Vert f\right\Vert _{L^{1}}\\
\left(\text{since }\theta\geq0\right) & \leq & \theta,
\end{eqnarray*}
a contradiction. Hence, $B\left(B_{1}\left(0\right)\right)\subset K_{2}$,
as desired.
