Of course can SR handle acceleration. Why shouldn't it?
I think the most simple example is the motion of a particle in an electromagnetic field, neglecting the radiation reaction. One should always start writing the mechanics in manifestly covariant form, using four-vectors. With the proper time ##\tau## of the particle you get
$$m \ddot{x}^{\mu}=K^{\mu},$$
where dots are derivatives wrt. proper time and ##K^{\mu}## is the socalled Minkowski force, which obviously is a four-vector since ##m## is a scalar (invariant mass of the particle) and ##x^{\mu}## is a four-vector and ##\tau## a scalar.
Now obviously
$$\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}. \qquad (*)$$
Taking the derivative of this equation wrt. to ##\tau## you get
$$\ddot{x}^{\mu} \dot{x}_{\mu}=0 \; \Rightarrow\; K^{\mu} \dot{x}_{\mu}=0.$$
The most simple equation of motion is thus given by using an antisymmetric 2nd-rank tensor field ##F_{\mu \nu}(x)## and writing
$$K^{\mu}=\frac{q}{c} F^{\mu \nu} \dot{x}_{\nu},$$
where ##q## is another scalar. This is indeed the Minkowski force for a particle in an electromagnetic field, which is given by the antisymmetric tensor field (which has 6 components, corresponding to the electric and magnetic field components of the electromagnetic field):
$$F^{j0}=E_j, \quad F^{jk}=-\epsilon^{jkl} B_l.$$
Here ##j,k,l \in \{1,2,3\}##.
Due to the given constraints following (*), you only need to consider the spatial part of the equation of motion. The time component then is automatically fulfilled. So let's write the spatial part of the equations first:
$$m \ddot{x}^j=\frac{q}{c} F^{j\nu} \dot{x}_{\nu}=\frac{q}{c} (F^{j0} c \dot{t}-F^{jk} \dot{x}_k) = \frac{q}{c} (E^j \gamma + \epsilon^{jkl} B_l \dot{x}_k)$$
or in vector notation
$$m \ddot{\vec{x}}=q \gamma\left (\vec{E} +\frac{\vec{v}}{c} \times \vec{B} \right).$$
Here I used ##\gamma=\mathrm{d} t/\mathrm{d} \tau=1/\sqrt{1-\vec{\beta}^2}## with ##\vec{\beta}=\vec{v}=\mathrm{d}_t \vec{x})##.
Writing ##\vec{p}=m \gamma \vec{v}## you get
$$m \dot{\vec{p}}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)$$
as expected.
You can check that the time component is automatically fulfilled and is just the work-energy theorem.