I am quite sure that Berezin integral cannot be interpreted as a sum in any sense. But one can find a geometric picture of Berezin integral.

**How should one think about supermanifold geometrically?** One way to think about manifold is following. It is a topological space and *sheaf of functions* (i.e. for each open subset one assign a ring). You may want to consider smooth, complex, algebraic etc manifold. Then those functions should be $C^{\infty}$, holomorphic or algebraic respectively.
(By the way, there are some conditions... you cannot take topological space to be a point and assign a ring of polynomials.)

For supermanifold rings (assigned to each open subspace) are supercommutative (i.e. $\mathbb{Z}/2\mathbb{Z}$ graded with $ab=(-1)^{|a| |b|} ba$). So the point is that if one factor over elements of degree 1, then it is an ordinary manifold. This manifold is called underlying manifold.

This picture recalls a standard picture from algebraic geometry about *non-reduced scheme*. Let me briefly remind you using an example. Consider intersection of circle $x^2 + y^2 = 1$ and a line $y=1$. This intersection is a point. On the other hand $\mathbb{k}[x,y]/(y-1, x^2+y^2-1) = \mathbb{k}[x]/(x^2)$. So the ring of functions is not $\mathbb{k}$ (as it should be for point). This is a double point. Also one can think about this double point as a fusion of two regular points.
To sum up, in ordinary algebraic geometry there may emerge nilpotents in a ring of functions. One should think about them as infinitesimal thickening of our initial variety.

The same happens with supermanifold. Note that all these odd coordinates are nilpotent. However, I would not recommend you to think about this odd direction as thickening. For example, correct notion of dimension of supermanifold is a number of even coordinates minus number of odd coordinates. I will try to justify this mysterious concept later.

*When I try to explain this concept to a physicist, I say that odd coordinates do not correspond to any physical direction. That is why Berezin integral is not a sum.*

**Construction of supermanifold** Let $M$ be a manifold an $E$ be a vector bundle over $M$. Let us define a supermanifold $\Pi E$. As topological space, it is homeomorphic to $M$. But the ring of functions is sections of $\Lambda^* (E^*)$. So locally there are $\dim M$ even coordinates and $\text{rank} E$ odd coordinates.

**Theorem** Any supermanifold is (not canonically) isomorphic to $\Pi E$ for some $E$.

**Canonical measure on $\Pi T M$**. The functions on this space is are differential forms $\theta_i = dx_i$. One can integrate a differential form of highest rank over $M$. We will see that this correspond to canonical measure.

One should think about $d \theta_i$ as a vector field since $\int \theta_j d \theta_i = \delta_{ij}$ (so $d \theta_i$ is dual basis to $\theta_i$). So $d \theta_1 \dots \theta_n dx_1 \dots dx_n$ is canonically defined.

Let $F(x, \theta)= f_0(x) + \dots + f_n(x) \theta_1 \dots \theta_n$
Then
$$\int_{\Pi TM} F(x, \theta) d \theta_1 \dots d\theta_n dx_1 \dots dx_n = \int_{M} f_n(x) dx_1 \dots dx_n$$
So this is standart integration of form of highest rank!
This picture justifies dimension formula since canonical measure can appear only on a zero-dimensional manifold (do not be too strict to this argument).

**Berezin measure**. Since any supermanifold is isomorphic to $\Pi E$, let us think about Berezin measure on $\Pi E$. It is section of $\Lambda E^* \otimes \Omega^{\dim M} M$. Integration is following procedure. First of all one use pairing $\Lambda^r E^* \otimes \Lambda^r E \rightarrow \mathbb{k}$. So Berezin measure defines a map

$$\Gamma ( \Lambda^r E ) \rightarrow \Omega^n M $$
Then one just integrate differential form.
This procedure explains geometric meaning of Berezin integral in terms of classical (not super) geometry.