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So this question states that a triangle is found where in between every $2$ numbers, their sum will be found in the next step, an example can be found below for the first $3$ steps. The question asks what will be the sum of the numbers in the $100^{text{th}}$ step? I managed to find a pattern that states $S_n=S_{n-1}+2 times 3^{n-1}$ if $S_k$ is the sum of step $k$. However, is this pattern correct and if it is then why? Also, can this triangle relate to Pascal's triangle? Thank you anyways.

Every number in the $n+1$-th row either comes directly from the $n$-th row, or is the sum of two neighbouring numbers in the $n$-th row. Every number in the $n$-th row has two neighbours (except the first and last) and so it is in two sums. So in summing the $n+1$-th row, we in fact sum every number from the $n$-th row three times (except the first and last). This yields the recursive formula
$$S_{n+1}=3S_n-2.$$

Your recurrence equation is also correct. Note that:
$$S_n=S_{n-1}+2cdot 3^{n-1}, S_1=4 Rightarrow frac{S_n}{3^n}=frac13cdot frac{S_{n-1}}{3^{n-1}}+frac23 Rightarrow \
a_n=frac13a_{n-1}+frac23, a_1=frac43 Rightarrow a_n=left(frac13right)^n+1=frac{S_n}{3^n} Rightarrow \
S_n=3^n+1.$$