I have never tried to keep straight the different sign conventions in the literature. There is a good reason for this: it is a theorem of category theory that any reasonable sign convention is as good as any other (viz: the monoidal category of $\mathbb Z/2$-graded vector spaces admits precisely two symmetric monoidal structures up to symmetric monoidal equivalence, determined by whether the non-unit invertible object is bosonic or fermionic); but it is therefore a corollary that whatever paper you're reading will use different sign conventions than you do, and therefore you should not trust their formulas (see J.P.Serre, How to write mathematics (badly)).
I bring this up to say that the difference between left and right differentiation, integration, etc., are ones purely of sign conventions.
Here's one way to develop the theory. Let $A$ be a supercommutative algebra, with multiplication $m: A \otimes A \to A$. The space $\mathrm{Der}(A)$ of left derivations of $A$ is the universal supervector space equipped with a map $\triangleright: \mathrm{Der}(A) \otimes A \to A$ equalizing the following two maps $\mathrm{Der}(A) \otimes A \otimes A \to A$:
$$ \triangleright \circ (\mathrm{id} \otimes m) \quad \text{and} \quad m \circ (\triangleright \otimes \mathrm{id}) + m\circ (\mathrm{id} \otimes \triangleright) \circ (\mathrm{flip} \otimes \mathrm{id})$$
Here $\mathrm{flip}: \mathrm{Der}(A) \otimes A \to A\otimes \mathrm{Der}(A)$ is the canonical commutivity isomorphism in the category of supervector space.
This definition makes the following clear. First, $\mathrm{Der}(A)$ is a kernel of two maps out of $\underline\hom(A,A)$ (enriched hom in the category of supervector spaces; i.e. $\underline\hom(X,-)$ is the right adjoint functor to $\otimes X$), and therefore exists and is a sub-supervector space of $\underline\hom(A,A)$.
Second, there is a clear corresponding notion of right derivation, and in fact a canonical isomorphism between the space of left derivations and the space of right derivations. Indeed, let $\mathrm{rightder}(A)$ denote the space of right derivations, so it comes with an action $\triangleleft : A\otimes \mathrm{rightder}(A) \to A$, and define $\triangleright: \mathrm{rightder}(A) \otimes A \to A$ by $\triangleright = \triangleleft \circ \mathrm{flip}$.
What the definition does not make particularly clear is how to calculate with derivations in practice.
Let $X$ be any supervector space. Henceforth, I will suppose that we work over a field of characteristic $0$. The supervector space $X^{\otimes n}$ has a canonical action of the symmetric group $S_n$, given on the generator $s_i = (i,i+1)$ by $\mathrm{id}_X^{\otimes (i-1)} \otimes \mathrm{flip} \otimes \mathrm{id}_X^{\otimes (n-i-1)}$. The quotient of this action is the supervector space $\mathrm{Sym}^n(X)$, and we define the supercommutative algebra $\mathrm{Sym}(X) = \bigoplus_n \mathrm{Sym}^n(X)$. Then just like in the usual case, there is a canonical isomorphism between $\underline\hom(X, \mathrm{Sym}(X))$ and $\mathrm{Der}(\mathrm{Sym}(X))$.
What many authors do is to work with underlying vector spaces. Let $\mathrm{Vect}$ and $\mathrm{SVect}$ denote respectively the categories of vector spaces and supervector spaces. There is a functor $u: \mathrm{SVect} \to \mathrm{Vect}$, the underlying vector space functor, given by sending both invertible objects to the one-dimensional vector space. It is monoidal, but not symmetric monoidal. This is why there is a mess of signs.
For example, you can choose an isomorphism $\hom(uX,u\mathrm{Sym}(X)) \cong u\hom(X,\mathrm{Sym}(X))$, and think of the latter as the space of left derivations. Then given $d \in \hom(uX,u\mathrm{Sym}(X))$, it defines naturally a left derivation (possibly odd). But it also defines a right derivation, and these will look different upon applying the functor $u$. Say, for example, that $X$ is the odd two-dimensional supervector space, and choose basis vectors $x,y\in uX$. What is the derivation corresponding to the map $uX \to u\mathrm{Sym}(X)$ sending $x \mapsto 1$ and $y\mapsto 0$? If we think left derivations are most natural, then this derivation might be
$$ \vec \partial_x : \begin{cases} 1 \mapsto 0 \\ x \mapsto 1 \\ y \mapsto 0 \\ xy \mapsto y \end{cases}$$
What I really mean is that this is the behavior seen after applying the functor $u$. If you instead favor right derivations, the formulas on underlying vectors look the same, except $xy = -yx \mapsto -y$. But, you know, I could have been perverse, and decided to work with a different, isomorphic functor $u'$, which had screwy sign conventions elsewhere. I mean, since $u$ is not symmetric, it has no a priori choice for how it interacts with things like $\mathrm{Sym}^n$; up to isomorphism the choice is made, but this doesn't force behavior on "basis vectors".
You also asked about integration. Let $X$ be a finite-dimensional totally-odd supervector space with dual space $X^* = \underline\hom(X,\mathbf 1)$, where $\mathbf 1$ denotes the ground field. There is a distinguished embedding $X^* \hookrightarrow \mathrm{Der}(\mathrm{Sym}(X))$ given by the identifications $\mathrm{Der}(\mathrm{Sym}(X)) \cong \underline\hom(X,\mathrm{Sym}(X))$ and $\mathrm{Sym}^0(X) \cong \mathbf 1$. Under the embedding into derivations, one gets an action $\triangleright|_{X^*} : X^* \otimes \mathrm{Sym}(X) \to \mathrm{Sym}(X)$.
It turns out that the cokernel of $\triangleright|_{X^*}$ is an invertible supervector space (either the even or the odd line). Which one it is depends on the parity of $\dim X$. The space of translation-invariant linear forms on $\mathrm{Sym}(X)$ is the dual space $(\mathrm{coker}(\triangleright|_{X^*}))^*$.
"The" Berezin integral is a basis vector for $u(\mathrm{coker}(\triangleright|_{X^*}))^*$. It is determined up to a scalar. A volume form on $uX$ determines "the" Berezin integral, but how it does so depends on your sign conventions, and I've never seen a good argument favoring one sign over the other. One convention I've seen is: if $uX$ is one-dimensional with basis vector $x$, then there is a Berezin integral $\int \mathrm d x$ (all one word) determined by $\int \mathrm d x(1) = 0$ and $\int \mathrm d x(x) = 1$. You can then extend this to higher-dimensional things, e.g. if $uX$ has basis $\{x,y\}$, then there is a Berezin integral $\int\mathrm d x \int\mathrm d y$, which satisfies for example
$$ \int\mathrm d x \int\mathrm d y(yx) = \int\mathrm d x \int\mathrm d y(y)(x) = \int\mathrm d x (1x) = 1. $$
And perhaps you decide that $\int\mathrm d x \int\mathrm d y = \iint \mathrm d x \mathrm d y$. But these are all choices you need to make.
Lastly, you asked about analogous of special functions. Let $X$ be totally even and finite-dimensional, and work over $\mathbb R$. You can define algebraically the Frechet topology on $\mathrm{Sym}(X)$, and its completion is the algebra of smooth functions on $X^*$. If $X$ is totally odd and finite-dimensional, then $\mathrm{Sym}(X)$ is finite-dimensional. Therefore all topologies are equivalent, and so "smooth" and "polynomial" and "power series" agree. As Ben McKay points out in the comments, you can certainly evaluate power series on odd variables.
The function of one even variable $x\mapsto \exp(\alpha x)$ is, up to a scalar, the unique eigenvector among smooth functions of $\partial_x$ with eigenvalue $\alpha$. You can therefore ask: in the odd case, does $\partial_x$ have any nonzero eigenvalues? The answer is no. So there is not an odd analogue of $\exp$.
