Part 1: Determining $f$ We let $f(0)=1$ and $y=x^2+x^3+x^4+\dotsb=\frac{x^2}{1-x}$. The generating function for $f$ is given by \begin{align*} \sum_{n\geq 0}f(n)x^n &= \sum_{k\geq 0}\sum_{\ell\geq 0}\binom{k+1+\ell}{\ell}x^ky^{\ell} \\ &=\sum_{k\geq 0}\frac{x^k}{(1-y)^{k+2}} \\ &= \frac{1}{(1-y)^2}\left(\frac{1}{1-\frac{x}{1-y}}\right) \\ &= \frac{(1-x)^2}{(1-x-x^2)(1-2x)} \\ &= \frac{1}{1-2x}-\frac{x}{1-x-x^2}.

From here we can see that $f(n)=2^n-F_n$ where $F_n$ is the $n$th Fibonacci number. Part 2: Calculating $\pmod{p}$ From the previous part, we wish to show that \[\sum_{1 \le i < j <k \le p-1 } \frac{F_i}{i \cdot j \cdot k} \equiv 0 \pmod{p}.\]This is as far as I've gotten. Perhaps someone else can do this part or I will revisit it later.

I'll complete a1267ab solution, using this very useful lemma and some congruences with multiplicative inverses. Lemma: For all $m,n\ge0$: $$\sum_{k=0}^{n} \binom{m+n}{k}(-1)^{n-k}F_{n-k}=-\sum_{k=0}^{n} \binom{m+k-1}{k}F_{n-k}$$Proof: Induction on $m+n$. For $m=n=0$: $$\sum_{k=0}^{n} \binom{m+n}{k}(-1)^{n-k}F_{n-k}=\binom{0}{0}(-1)^{0}F_{0}=0=-\binom{0+0-1}{0}F_{0}=-\sum_{k=0}^{n} \binom{m+k-1}{k}F_{n-k}$$For $m=0,n=1$: $$\sum_{k=0}^{n} \binom{m+n}{k}(-1)^{n-k}F_{n-k}=\binom{1}{0}(-1)^{1}F_{1}+\binom{1}{1}(-1)^{0}F_{0}=-1=-\binom{0+0-1}{0}F_{1}-\binom{0+1-1}{1}F_{0}=-\sum_{k=0}^{n} \binom{m+k-1}{k}F_{n-k}$$For $m=1,n=0$: $$\sum_{k=0}^{n} \binom{m+n}{k}(-1)^{n-k}F_{n-k}=\binom{1}{0}(-1)^{0}F_{0}=0=-\binom{1+0-1}{0}F_{0}=-\sum_{k=0}^{n} \binom{m+k-1}{k}F_{n-k}$$Now, suppose that for, $m+n\le a+b$, the lemma is true. We have, by hypothesis: $$\sum_{k=0}^{b} \binom{a+b}{k}(-1)^{b-k}F_{b-k}=-\sum_{k=0}^{b} \binom{a+k-1}{k}F_{b-k} (m=a,n=b)...(1)$$$$\sum_{k=0}^{b+1} \binom{a+b}{k}(-1)^{b+1-k}F_{b+1-k}=-\sum_{k=0}^{b+1} \binom{a+k-2}{k}F_{b+1-k} (m=a-1,n=b+1)$$$$\Rightarrow (-1)^{b+1}F_{b+1}+\sum_{k=0}^{b} \binom{a+b}{k+1}(-1)^{b-k}F_{b-k}=-\left[ F_{b+1} + \sum_{k=0}^{b} \binom{a+k-1}{k+1}F_{b-k} \right]...(2)$$$(1)+(2)$ yelds: $$\Rightarrow (-1)^{b+1}F_{b+1}+\sum_{k=0}^{b} \left(\binom{a+b}{k}(-1)^{b-k}F_{b-k} + \binom{a+b}{k+1}(-1)^{b-k}F_{b-k} \right)=-\left[ F_{b+1} + \sum_{k=0}^{b} \left(\binom{a+k-1}{k}F_{b-k} + \binom{a+k-1}{k+1}F_{b-k} \right) \right]$$$$\Rightarrow \binom{a+b+1}{0}(-1)^{b+1}F_{b+1}+\sum_{k=0}^{b} \binom{a+b+1}{k+1}(-1)^{b-k}F_{b-k} =-\left[ \binom{a-1}{0}F_{b+1} + \sum_{k=0}^{b} \binom{a+k}{k+1}F_{b-k} \right]$$$$\Rightarrow \sum_{k=0}^{b+1} \binom{a+b+1}{k+1}(-1)^{b+1-k}F_{b+1-k} =-\left[ \sum_{k=0}^{b+1} \binom{a+k-1}{k}F_{b+1-k} \right]$$ and the lemma follows by induction. Put $m=0, n=p$ on the lemma. We have: $$\sum_{k=1}^{p-1} \binom{p}{k}(-1)^{p-k}F_{p-k}=0 ...(*)$$Put $m=p, n=p$ on the lemma. We have: $$\sum_{k=0}^{p} \binom{2p}{k}(-1)^{p-k}F_{p-k}= -\sum_{k=0}^{p} \binom{p+k-1}{k}F_{p-k} ...(**)$$ Let $H_{k-1} = \sum_{0<i<k} \frac{1}{i}$ and $L_{k-1}=\sum_{0<i<j<k} \frac{1}{ij}$. Then: $$\binom{p}{k}=\dfrac{p}{k} \binom{p-1}{k-1} = (-1)^{k-1}.\dfrac{p}{k} \left( \prod_{0<j<k} \left( 1-\dfrac{p}{j} \right) \right)$$$$\Rightarrow \binom{p}{k} \equiv (-1)^{k-1}. \dfrac{p}{k}.\left( 1 - pH_{k-1} +p^2 L_{k-1} \right) \pmod{p^4}...(a)$$In a similar way, we have: $$\binom{2p}{k} \equiv (-1)^{k-1}. \dfrac{2p}{k}.\left( 1 - 2pH_{k-1} +4p^2 L_{k-1} \right) \pmod{p^4}...(b)$$$$\binom{p+k-1}{k} \equiv \dfrac{p}{k}.\left( 1 + pH_{k-1} + p^2 L_{k-1} \right) \pmod{p^4}...(c)$$Substituting $(a),(b),(c)$ in $(*),(**)$, we have: $$(*): pS_1-p^2 S_2 +p^3 S_3 \equiv 0 \pmod{p^4} $$$$(**): 2pS_1 -4p^2 S_2 +8p^3 S_3 \equiv -pS-1 -p^2 S_2 -p^3 S_3 \pmod{p^4} $$Where: $$S_1 = \sum_{0<i<p} \dfrac{F_{p-i}}{i};S_2 = \sum_{0<i<j<p} \dfrac{F_{p-j}}{ij};S_3 = \sum_{0<i<j<k<p} \dfrac{F_{p-k}}{ijk}$$Then, $pS_1-p^2 S_2 \equiv -p^3 S_3 \pmod{p^4} $ and $3pS_1-3p^2 S_2 \equiv -9p^3 S_3 \pmod{p^4}$, so $6p^3 S_3 \equiv 0 \pmod{p^4}$, and since $p>3$, we have $p|S_3$, as desired.

Here's another way of determining $f$ (I think this one is closer to the proposer's approach) $$ f(n)=\sum\binom{1+t_1+t_2+\dots+t_n}{1+t_1,t_2,\dots,t_n} $$Let $t_1'=1+t_1\ge 1$ , so we only need to find all n-tuples $(t_1',t_2,\dots,t_n)$ that satsisfies $t_1'+\sum_{j>1}j\times t_j=n+1,t_1'\ge 1,t_2,\dots,t_n\ge 0$ . Notice that this is actually the number of solution with $t_1'\ge 0$ subtracted by the number of solution that $t_1'=0$ . By generating function we have $$ \begin{aligned} \sum f(n)x^{n+1}&=\sum (x+x^2+\cdots)^k-\sum (x^2+x^3+\cdots)^k\\ &=\sum(\frac{x}{1-x})^k-\sum(\frac{x^2}{1-x})^k\\ &=(1-\frac{x}{1-x})^{-1}-(1-\frac{x^2}{1-x})^{-1}\\ &=\frac{1-x}{1-2x}-\frac{1-x}{1-x-x^2}\\ &=(1+\frac{x}{1-2x})-(1+\frac{x^2}{1-x-x^2})\\ &=x(\frac{1}{1-2x}-\frac{x}{1-x-x^2})\\ \end{aligned} $$