What is the coefficient of $x^{9}$ in the expansion of $(1+x)^{12}(1+x)^{4}$?

You've stumbled across a case of the Vandermonde identity:

$$ \sum_{k=0}^r{m\choose k}{n\choose r-k} = {m + n\choose r} $$

So you are indeed correct, the answer is equivalent to the coefficient of $x^9$ in $(1+x)^{16}$, since $$ {16\choose 9} = \sum_{k=0}^9{4\choose k}{12\choose 9-k} $$ To obtain the second form, note that $$ \begin{aligned} \ [x^9](1+x)^{12}(1+x)^4 & = [x^9]\left(\sum_{k=0}^{12}{12\choose k}x^k\right)\left(\sum_{k=0}^4{4\choose k}x^k\right) \\ & = [x^9]\sum_{k=0}^{12+4}x^k\left(\sum_{i,j\ge0,\ i+j=k}{4\choose i}{12\choose j}\right) \\ & = [x^9]\sum_{k=0}^{16}x^k\left(\sum_{j=1}^{k}{4\choose j}{12\choose k-j}\right) \\ & = \sum_{j=0}^9{4\choose k}{12\choose 9-k} \end{aligned} $$

Going from the first line to the second line, we've essentially used a finite version of the Cauchy Product, and to go from the inner sum on the second line to the inner sum on the third line, note that if $i+j=k$, then $i=k-j$, so really we only care about $j$.

If you think about this as choosing objects, the result is actually very intuitive. Picture twelve copies of $(1+x)$ laid out. The only way to get a term of $x^9$ when we expand is to pick exactly $9$ $x$'s and $7$ $1$'s. Alternatively, if we have $12$ copies of $(1+x)$ followed by $4$ more, we can imagine picking the first $k$ $x$'s from the first $12$, and the remaining $9-k$ from the next $4$.