Currents are distributions dual to smooth forms. I often get confused about the degrees, so I'm never sure what the standard meaning of an $m$-current. But according to Smirnov (Sec 1.4) an $m$-current is an element of the dual space of smooth $m$-forms. As is usual with distributions, differential operators on forms can be extended to differential operators on currents via the adjoint formula. So, one can define a differential $\partial$ on currents as the adjoint of the de Rham differential $d$ on forms: $\langle \partial T, \alpha \rangle = \langle T, d\alpha \rangle$, up to a choice of sign.
One can embed smooth $(n-m)$-forms in $m$-currents (on an $n$-dimensional space) using the formula $\langle T_\alpha, \beta \rangle = \int \alpha \wedge \beta$. One must restrict the support so that this integral is always defined. That is, if $\beta$ is allowed to be arbitrary, then $\alpha$ must have compact support, and vice versa. Restricted to the embedded form, the $\partial$ operator is just the de Rham operator $d$ (up to sign). This follows from the formula $\int (d \alpha) \wedge \beta = \pm \int \alpha \wedge (d\beta)$.
Since it follows from the definition that $\partial^2 = 0$, one can use $\partial$ to define a cohomology on currents. Turns out that this cohomology on currents is isomorphic to the de Rham cohomology of the embedded forms. So, the cohomology on $m$-currents with compact support is isomorphic to the de Rham cohomology with compact support in degree $n-m$.
So, if I interpret your notation correctly, the claim that for an $(n-1)$-current $T$, the identity $\partial T = 0$ implies $T = \partial S$ for some $n$-current $S$, is simply statement that the de Rham cohomology of $1$-forms (with appropriate support) vanishes. Since, I think, Smirnov is working on $\mathbb{R}^n$, that is only to be expected. If you restrict $T$ and $S$ to be embedded smooth forms, then $T$ is simply a $1$-form and $S$ a $0$-form, with $d S = T$, which is consistent with the interpretation proposed by Smirnov.
I'm not sure that I can help with showing, under relaxed smoothness conditions, that $S$ necessarily belongs to the BV class.
BTW, the details of the above discussion can be found in de Rham's own book. Perhaps also in the Federer reference given by Smirnov.
Update: Above, I gave a formula for embedding smooth $(n-m)$-forms in $m$-currents. If you have a preferred measure $\mu$ on your space (like the Lebesgue measure on $\mathbb{R}^n$, for instance), then you can also embed $m$-polyvectors in $m$-currents. In coordinates, $m$-polyvectors have $m$ contravariant indices, which are fully antisymmetriezed. So a $1$-polyvector is just a vector field. The embedding in currents uses the formula $\langle T_v, \alpha \rangle = \int (v\cdot \alpha) d\mu$, where $\cdot$ denotes the contraction of the $m$ contravariant indices of $v$ and the $m$ covariant indices of $\alpha$. If you work out the adjoint formula for the differential acting on an embedded vector field $v$, you find that $\partial T_v = T_{\operatorname{div} v}$ (up to sign), where in Cartesian coordinates $\operatorname{div} v = \partial_i v^i$.
Note that the above formulas for embedding forms and vector fields and currents are different. In particular, if you use the metric to lower indices and thus identify vector fields and $1$-forms, you also get two different ways of embedding vector fields (and polyvector fields) in currents. So if you start with a vector field $v$ ($1$-polyvector), then you can either embedded it in $(n-1)$-currents (first method above) or in $1$-currents (second method above). You get different actions of $\partial$ on $v$ using different embeddings. I already gave the formula for the second embedding, for the first embedding you find $(\partial v)^{ij} = \partial^{[i}v^{j]}$ (up to sign, and all in Cartesian coordinates). You can see that this is the formula for the exterior differential of forms, just with all indices raised. I believe that it is this last formula that Smirnov has in mind when talking about $(n-1)$-currents.
