Is there an analogous version of $\operatorname{var(X+cY)} = \operatorname{var}(X) + c^2\operatorname{var}(Y)$ for vectors of uncorrelated random variables ($X$ and $Y$ are random variables here)?
For example, consider $\operatorname{var(\boldsymbol{X}+A\boldsymbol{var}(Y)$, where the bold indicators a vector random variables. Would the variance look something like
$$ \operatorname{var(\boldsymbol{X})+A^TA\operatorname{var}(\boldsymbol{Y})} $$ ?
For vector-valued random variables $\overrightarrow{X}$ and $\overrightarrow{Y}$, the appropriate concept is a covariance matrix: if $\overrightarrow{X} = [x_1, ..., x_n]$, then
$$\operatorname{cov}(\overrightarrow{X}) = [a_{ij}],$$
where $a_{ij} = \operatorname{cov}(x_i, x_j)$. It is then true that for any matrix $A$ of appropriate dimension,
$$\operatorname{cov}(A\overrightarrow{X}) = A\operatorname{cov}(\overrightarrow{X})A^t,$$
and direct computation should show that $\operatorname{cov}(\overrightarrow{X} + \overrightarrow{Y}) = \operatorname{cov}(\overrightarrow{X}) + \operatorname{cov}(\overrightarrow{Y})$ when the components of $\overrightarrow{X}, \overrightarrow{Y}$ are uncorrelated. So yes, you do get a version of your original statement:
$$\operatorname{cov}(\overrightarrow{X} + A\overrightarrow{Y}) = \operatorname{cov}(\overrightarrow{X}) + A\operatorname{cov}(\overrightarrow{Y})A^t,$$
and if $A = cI$ for some scalar $c$, then it is even true that
$$\operatorname{cov}(\overrightarrow{X} + c\overrightarrow{Y}) = \operatorname{cov}(\overrightarrow{X}) + c^2\operatorname{cov}(\overrightarrow{Y}).$$