This recent question got me thinking, if a textbook (or an exam) tells a student to give a combinatorial proof of something involving (sums of) binomial coefficients, would it be enough to show that by Pascal's triangle these things do add up, or would you fail an answer like that? What if we didn't call it Pascal's triangle but "the number of paths that stop at some point at step $i$ during a one-dimensional random walk"?
[Math] Pascal’s triangle and combinatorial proofs
combinatoricssoft-question
Related Solutions
When $p = 2$, Lucas' Theorem states that ${n \choose m} \equiv 0 \pmod{2}$ if and only if in the binary expansion of $n = (\overline{ \cdots n_1n_0})_2$ and $m = (\overline{ \cdots m_1m_0})_2$, some binary digit (say the $d$th binary digit) satisfies $n_d = 0, m_d = 1$. In particular, this shows that ${2^k + n \choose m} \equiv {2^k + n \choose 2^k + m} \equiv {n \choose m} \pmod{2}$ when $2^k > n > m$, which describes the recursive nature of the Sierpinski triangle that can be found in Pascal's triangle by highlighting the odd elements (which it seems is the geometric interpretation you referred to). However, this is a fairly restricted application of Lucas' Theorem and can be seen by easier means.
Here's the full combinatorial argument. Raney's lemma was about half of it, I'd say. The argument shows that $S(n,2) = \binom{3n-1}{n} \frac{3}{3n-1}$ (which is equivalent to the two formulations I give in the question). At the end I'll discuss the generalization to the $r$ case.
Intro.
Consider the sequences with $2n$ occurrences of $-1$ (i.e., $2n$ heads) and $n$ occurrences of $+2$ (i.e., $n$ tails). We want to show that the number of these sequences with all partial sums nonzero is $\binom{3n-1}{n} \frac{3}{3n-1}$. The complete sum and empty sum are clearly $0$, so "partial sum" excludes those two cases. The sequences we want to count can be split into three groups: (1) all partial sums positive, (2) all partial sums negative, (3) some partial sums positive and some negative.
Group 1: The number of these sequences with all partial sums positive is $\binom{3n-1}{n} \frac{1}{3n-1}$.
This is the part that uses Raney's lemma. If all partial sums are positive, the last element in the sequence must be $-1$. Thus we want to count the number of sequences with $2n-1$ occurrences of $-1$ and $n$ occurrences of $+2$ that add to $+1$ and have all partial sums positive. Ignoring the partial sums restriction, there are $\binom{3n-1}{n}$ such sequences. If we partition these $\binom{3n-1}{n}$ sequences into equivalence classes based on cyclic shifts, Raney's lemma says that exactly one sequence in each equivalence class has all partial sums positive. Because there are $3n-1$ elements in each sequence there $3n-1$ sequences with the same set of cyclic shifts, and so there are $3n-1$ sequences in each equivalence class. Thus the number of sequences in Group 1 is $\binom{3n-1}{n} \frac{1}{3n-1}$.
Group 2: The number of these sequences with all partial sums negative is also $\binom{3n-1}{n} \frac{1}{3n-1}$.
To see this, just reverse the sequences counted in Part 1.
Group 3: The number of these sequences with some positive partial sums and some negative partial sums is, yet again, $\binom{3n-1}{n} \frac{1}{3n-1}$.
This one is a little trickier. First, because of the $-1$'s, it is not possible to switch from positive partial sums to negative partial sums. Thus any sequence counted here must have exactly one switch: from negative partial sums to positive partial sums. The switch must occur at some point where the partial sum is $-1$ and the next element is $+2$. Thus we have some sequence $(a_1, \ldots, a_m, +2, a_{m+2}, \ldots, a_n)$ where the sums $a_1, a_1 + a_2, \ldots, a_1 + \ldots + a_m$ are all negative. Consider the sequence $(+2, a_m, \ldots, a_2, a_1, a_{m+2}, \ldots, a_n)$. Since $+2 + a_m + \ldots + a_1 = 1$, and $a_k + a_{k-1} + \ldots + a_1 < 0$ for all $k$, $1 \leq k \leq m$, it must be the case that $+2 + a_m + \cdots + a_{k+1} > 1$ for all $k$, $1 \leq k \leq m-1$. So the sequence $(+2, a_m, \ldots, a_2, a_1, a_{m+2}, \ldots, a_n)$ is in Group 1. To see that this mapping is a bijection, note that any sequence in Group 1 must start with $+2$ and have a first time that a partial sum is equal to $+1$. Thus this transformation is reversible.
Summing up.
Putting Groups 1, 2, and 3 together we see that the total number of sequences we want to count is $\binom{3n-1}{n} \frac{3}{3n-1}$.
Generalization to the $r$ case.
The arguments for Group 1 and Group 2 adapt in a straightforward manner for general $r$; we get $\binom{(r+1)n-1}{n} \frac{1}{(r+1)n-1}$ for both. Sequences in Group 3 can still only switch once - from negative partial sums to positive partial sums. But Group 3 can be broken up into subgroups depending on whether the first partial sum to become positive is $+1, +2, \ldots, r-1$. Using transformations like the one described above for Group 3 in the $r = 2$ case, we can show that there is a bijection between each of these $r-1$ subgroups and Group 1. Thus there are $\binom{(r+1)n-1}{n} \frac{r-1}{(r+1)n-1}$ sequences in Group 3. In total, then, $S(n,r) = \binom{(r+1)n-1}{n} \frac{r+1}{(r+1)n-1}$.
Final comment.
All of this is may be easier to visualize in terms of paths from $(0,0)$ to $((r+1)n,(r+1)n)$ that use right steps of size $1$ and up steps of size $r$ and that do not step on the diagonal except at $(0,0)$ and $((r+1)n,(r+1)n)$. (They can cross the diagonal, though.) I found it easier to discuss the partial sums interpretation, though, given the difficulty of creating such graphics on this forum.
Best Answer
In the context of your question, talking about textbooks or exams, this could be viewed as a question of pedagogy. There are a couple of reasons why an author or an instructor would introduce combinatorial proofs, particularly those involving binomial coefficients:
A good example of the latter is the well-known identity $$ \sum_{k=0}^n\binom{n}{k}=2^n $$ One could do this by induction on $n$, but the algebraic proof is nowhere as simple (and illustrative) as the purely combinatorial version.
To answer your question, then, I'd recommend one of two strategies. First, if this is a course and you were posed such a question, the best way would be to ask your instructor something like "I could answer this question by using properties of Pascal's Triangle; would you accept that or do you want a combinatorial proof from the ground up?" If such a question would be considered out of bounds or if you were self-studying, you might ask yourself, "What was the author's likely intent here, to have me get the answer by any means possible or was it to encourage me to think combinatorially?"