I am looking for (family of) real-valued functions $f:\mathbb{N}^k \rightarrow \mathbb{R}$ such that for any $i$ one can compute $f(n_1, n_2, \dotsc,n_i+1, \dotsc, n_k)$ knowing only $f(n_1, n_2, \dotsc,n_i, \dotsc, n_k)$.
More rigorous description of the problem. For a fixed $k$ find a function $f:\mathbb{N}^k \rightarrow \mathbb{R}$ and $k$ functions (which could be the same) $\phi_1, \phi_2, \dotsc, \phi_k : \mathbb{R} \rightarrow \mathbb{R}$, satisfying: $$ f(n_1, n_2, \dotsc,n_i+1, \dotsc, n_k) \equiv \phi_i \left( f(n_1, n_2, \dotsc,n_i, \dotsc, n_k) \right) \text{ for all } n_1, n_2, \dotsc, n_k \in \mathbb{N} \,. $$
The obvious solution would be just a sum of arguments: $$f(n_1, n_2, \dotsc, n_k) = n_1 + n_2 + \dots + n_k \,,$$ $$\phi_i(x) = x+1 \text{ for all } i \,.$$
Indeed, $f(n_1, n_2, \dotsc,n_i+1, \dotsc, n_k) = f(n_1, n_2, \dotsc,n_i, \dotsc, n_k) + 1$.
I cannot come up with any more sophisticated examples of such functions
Let $f:\mathbb{N}^k \rightarrow \mathbb{R}$ be an injective function. Then there is a function such $g:\mathbb{R} \rightarrow \mathbb{R}^k$ such that for all $X\in \mathbb{N}^k$, $g(f(X))=X$. For any $1 \leq i \leq k$ define the functions $\phi_i: \mathbb{R} \rightarrow \mathbb{R}$ as follow $$\phi_i(t)=f(g(t)+E_i)$$ for all $t\in f(\mathbb{N}^k)$ and else $\phi_i(t)=0$, in which for any $1 \leq i \leq k$, $E_i=(a_1,...,a_k)$ such that $a_i=1$ and $a_j=0$ for any $j\neq i$. Now for any $X=(a_1,...,a_k)\in \mathbb{N}^k$: $$\phi_i(f(n_1, n_2, \dotsc,n_i+1, \dotsc, n_k))=\phi_i(f(X))=f(g(f(X))+E_i)=f(X+E_i)=f(n_1, n_2, \dotsc,n_i+1, \dotsc, n_k).$$ Therefore for any real-valued injective functions $f:\mathbb{N}^k \rightarrow \mathbb{R}$ satisfy in your question.
Hint. If $f:\mathbb{N}^k \rightarrow \mathbb{R}$ does not an injective function, but $f(\mathbb{N}^k)$ is a countable set, so with axiom of choice you can build functions $\phi_i$ that satisfy your condition.
Proof. Let $h_i:\mathbb{R} \rightarrow \mathbb{R}$ are functions such that $f(X+E_i)-f(X)=h_i(f(X))$ for a fixed real $c$ and any $1 \leq i \leq k$ and $X\in \mathbb{N}^k$. Then we can define $\phi_i(t)=h_i(t)+t$ for any $1 \leq i \leq k$.
Example. Let $f(n_1, n_2, \dotsc, n_k)=n_1n_2\dots n_k$, then we have $$f(X+E_i)-f(X)=\frac{n_1n_2\dots n_k}{n_i}=\frac{1}{n_i}f(X)=h_i(f(X))$$ therefore we get that $$\phi_i(t)=(\frac{1}{n_i}+1)t.$$