Let $X$, $Y$, and $Z$ be independent $U(0,1)$ random variables, and set $X_1=X+Z$ and $Y_1=Y+2Z$.
The question is asking for $Var(X_1 + Y_1)$, and the provided solution uses the covariance since $X_1$ and $Y_1$ are dependent.
I was wondering if it was possible to get $Var(X_1 + Y_1)$ without using the covariance. I used these steps: \begin{align} Var(X_1+Y_1) &= Var(X+Z + Y + 2Z) \\ &= Var(X + Y + 3Z) \\ &= Var(X) + Var(Y) + Var(3Z) \\ &= Var(X) + Var(Y) + 9 * Var(Z) \\ &= \frac{1}{12} + \frac{1}{12} + 9 * \frac{1}{12} \\ &= \boxed{\frac{11}{12}} \end{align} The answer is the same as the provided solution's answer. Are the steps valid?
Your approach is entirely valid. Of course, it is consistent with using the bilinearity of covariance.
$\begin{align} &\quad\mathsf{Var}(X_1+X_2) \\ &=\mathsf{Var}(X_1)+\mathsf{Var}(X_2)+2\mathsf{Cov}(X_1,X_2) \\ &= \mathsf{Var}(X+Z)+\mathsf{Var}(Y+2Z)+2\mathsf{Cov}(X+Z,Y+2Z) \\ &= \mathsf{Var}(X)+\mathsf{Var}(Z)+\mathsf{Var}(Y)+4\mathsf{Var}(Z)+2\big(\mathsf{Cov}(X,Y)+2\mathsf{Cov}(X,Z)+\mathsf{Cov}(Z,Y)+2\mathsf{Var}(Z)\big) \\ &= \mathsf{Var}(X)+\mathsf{Var}(Y)+9\mathsf{Var}(Z)+0 \end{align}$
I will say that your approach is clearer.
$\begin{align} &\quad\mathsf{Var}(X_1+X_2) \\ &= \mathsf{Var}(X+Z+Y+2Z) \\ &= \mathsf{Var}(X)+\mathsf{Var}(Y)+\mathsf{Var}(3Z) \\ &= \mathsf{Var}(X)+\mathsf{Var}(Y)+9\mathsf{Var}(Z) \end{align}$
It comes down to $\mathsf{Var}(aZ+bZ)\\=\mathsf {Var}(aZ)+\mathsf {Var}(bZ)+2\mathsf{Cov}(aZ,bZ)\\=(a^2+2ab+b^2)\mathsf{Var}(Z)\\=(a+b)^2\mathsf{Var}(Z)\\=\mathsf {Var}((a+b)Z)$