Here is an experimental answer I obtained, via Mathematica. It seems something quite subtle
may be going on--or maybe not.

The two functions f I am interested in were given in the question.

The first function $v=O(n/\log n)$ was tested in Mathematica as follows, e.g.,

$g[c_,d_,n_]:= c*n/Log[n]^d;$

$L1=Table[Min[DivisorSigma[0,n],g[0.5,1,10^5]],{n,2,10^4}];$

$L2=Accumulate[L1]; DiscretePlot[L2[k]/(k*Log[k]),{k,2,10^4}]$

Note that DivisorSigma[0,n] is the function that sums 0th powers of divisors of n.
I am comparing the accumulated truncated divisor sum to the well known approximation in
the first equation for non-truncated divisor sums.

The ratio, as expected, was essentially 1 with very little variation, for all sorts of values
of $[c,d] \in [0,1]\times [0,1]$, some of which were as follows:
$$[0.5,1],[0.05,1], [0.5, 0.5], [0.5, 0.05]$$

The surprising results were obtained for the second function $v=O(n^{\alpha})$.

$h[a_,b_,n_]:= a*n^b;$

$L1=Table[Min[DivisorSigma[0,n],h[1,0.5,10^5]],{n,2,10^4}];$

$L2=Accumulate[L1]; DiscretePlot[L2[k]/(k*Log[k]),{k,2,10^4}]$

In this case, very unexpectedly, the ratio in the plot was sometimes tending to values
way below 1. I again tested for values
of $[a,b]$, some of which were as follows:

$$[1,0.75],[1,0.5], [1,0.25], [1, 0.05],[0.5,1],[0.05,1],[0.25,0.5],[0.25,0.75],[0.05,0.75]$$

Again, if $b=1$ which is the exponent of $n$ in $v$ is 1, the ratio is trivially 1.

However, for example, $b=0.75$, with $a=0.05$ or larger still gives the ratio 1, all accross the
plot for $k=2,\ldots,10^4$.

As for $b=0.5$ it gives 1 for $a=1$ but not for $a=0.25$ where it seems to tend to 0.95 but maybe it will shoot back up to 1 for extremely large values of $n$ which I can't test.

The same effect is more pronounced, for $b=0.25,$ where even $a=1$ can't stop the plot from going below 0.8 and for $b=0.05$ where it goes below 0.2 for $a=1$ and below 0.1 for $a=0.5$

I think that all this may be just finite effects and the answer for both truncation functions $v$ is that the sum of $\min [d(k),v(n)]$ over $1\leq k\leq n$ is $O(n \log n)$.

firstsum, i.e., is asymptotically $\sum_{n\leq x} d(n)$, for $x$ large enough. – kodlu Apr 3 '12 at 23:56