As I understand the claim $\Lambda(x,n) = \frac{n'}{x'} \Phi(x) \pm V$, it is false for some $n$ and $x$ with $n$ close to $x$. Let us take $x$ to be $P_4=210$, the fourth primorial. Let us take $n$ close to and less than $210$, say $208$. (I use Gerry's observation that $n'/x' = n/x$.)
The claim says that the number of totatives of $210$ which are less than or equal $208$ is close to the expected number, with an error $V$ bounded by $\frac{x' -n'}{x'n'}$ times the expected value. For $x=210$ the expected value is $48$, and so the number of totatives
given by the formula is in the range (my interpretation of $=$)
$(208/210)*48 \pm 48*(2/(208*210))$, which I rewrite as $48 - 16/35 \pm 1/455$. (See comment below for the missing factor.) The allowed range does not even contain an integer, much less the correct answer which is 47. For the same $x$, one can find further $n$ which violate the claim.

One should note that $\Lambda(x,n) + \Lambda(x, x-n) = \Phi(x)$, and that the error for an interval of length $n$ should be the same (up to sign) as the error for length $x-n$. The current estimate does not show such a symmetry.

In an earlier question I referenced D. Lehmer's 1955 work on "Distribution of Totatives". One of his results there I can provide here with a hint of proof. Take a large squarefree integer $x$ with a prime factor $p$ of $x$, and assume $x \gt p$. The range of totatives of $x$ in $(0,x)$ is like $p$ copies of the range of totatives of $n= x/p$ concatenated together, with one copy dilated by $p$ subtracted off. If you look at the interval $(0,x/(p-1))$ and count totatives of $x$, you get $p/(p-1)$ many totatives of $n$ [actually, the totatives of $n$ in the interval $(0,n(p/(p-1)))$ ] minus $1/(p-1)$ of the scaled copy of totatives of $n$, which is in 1-1 correspondence with the totatives of n in
$(n,n(p/(p-1)))$. So the number of totatives of $x$ in $(0, x/(p-1))$ is the same as the number of totatives of $n$ in $(0,n)$, which is $1/(p-1)$ times the number of totatives of $x$ in $(0,x)$.

Thus one of Lehmer's results in the article implies that for any prime factor $p$ of $x$, the interval of length $x/(p-1)$ beginning at 0 contains exactly the expected number of totatives. Further, this argument holds for other integers $j$ and intervals $(jx/(p-1), (j+1)x/(p-1))$. So there are some intervals for which it can be easily shown that $V=0$. If $x$ is divisible by the square of a prime $p$, it is easy to see that $(jx/p, (j+1)x/p)$ contains the expected number of totatives as well.

For more general intervals that do not have $x/p$ or $x/(p-1)$ or some multiple of such as endpoints, one can still get some modest results. As pointed to in this question, if $x$ is squarefree with $k$ distinct prime factors, an interval of the form $(0,n]$ has as many totatives of $x$ as the expected number plus or minus an error bounded by $2^{k-1}$. We can concatenate errors to say the same for an interval $(m,n]$, but with error bounded by $2^k$.

I have not seen published anywhere the following, and would appreciate references. Let $x$ be squarefree with $k \gt 2$ prime factors, and $p,q$ distinct primes dividing $x$. Let $n$ be an arbitrary real. The number of totatives of $x$ in the interval $(n, n+L]$ is different from the expected number by at most a) $2^{k-1}$ if $L= x/p$ or $x/q$, b) $3(2^{k-2})$ if $L=x/(pq)$. (One can extend this to larger sets of primes and smaller versions of $L$, but not with satisfactory results.) Unfortunately this does not mean that smaller length intervals yield smaller deviations from the expected value.

Gerhard "Been Thinking Lots About This" Paseman, 2015.09.23