Actually, the two procedures are the same. The difference between $$ \Delta Y_t = B\Delta X_t + \Delta \epsilon_t $$ and $$ \Delta Y_t = B\Delta X_t + v_t $$ is that you can estimate the second but not the first because you don't observe $\epsilon_t$. So the first equation is rather a theoretical model whilst the second is the estimating equation that you would use in practice. If you wanted to directly subtract $Y_{t-1}$ from both sides manually then this can only be done if you observe the true errors. You will notice that $v_t$ is an estimate of $\epsilon_t$. Re-arrange the theoretical model and the regression equation, if $\Delta Y_t - B\Delta X_t = \Delta \epsilon_t$ and $\Delta Y_t - B\Delta X_t = v_t$, then it must be true that $\Delta \epsilon_t = v_t$. Consider a simple example with two time periods and $B=0.3$ being constant over time.
$$ \begin{array}{c|lc|r} time & Y_t & X_t & Y_t - BX_t =v_t \\ \hline 1 & 10 & 17 & 10-0.3\cdot17 =4.9 \\ 2 & 13 & 21 & 13 - 0.3\cdot 21 = 6.7 \\ \hline \Delta & 3 & 4 & 3 - 0.3\cdot 4 = 1.8 \end{array} $$$$ \begin{array}{c|lc|r} time & Y_t & X_t & Y_t - BX_t =v_t \\ \hline 1 & 10 & 17 & \\ 2 & 13 & 21 & \\ \hline \Delta & 3 & 4 & 3 - 0.3\cdot 4 = 1.8 \end{array} $$
Suppose that $v_t$ was a consistent estimate of $\epsilon_t$ in all periods (which is true here because we have deterministically specified the data generating process by fixing $B$), then $\epsilon_2 - \epsilon_1 = \Delta \epsilon_t = 4.9 - 6.7 = 1.8$ which$\widehat{v}_t = \Delta \epsilon_t = 1.8$ is the same as the residual from our second regression as an estimate of the error of the first equation.