Variance over IID a random number of times
The law of total variance says $$ \operatorname{var}(Y) = \operatorname{E}(\operatorname{var}(Y\mid X)) + \operatorname{var}(\operatorname{E}(Y\mid X)). $$ So $$ \begin{align} \operatorname{var}\left(\sum_{i=1}^N X_i\right) & = \operatorname{E} \left(\operatorname{var}\left(\sum_{i=1}^N X_i \mid N\right)\right) + \operatorname{var}\left(\operatorname{E} \left(\sum_{i=1}^N X_i \mid N\right)\right) \\ \\ \\ & = \operatorname{E}(N\operatorname{var}(X)) + \operatorname{var}(N\operatorname{E}(X)) \\ \\ & = \operatorname{var}(X)\operatorname{E}(N) + (\operatorname{E}(X))^2 \operatorname{var}(N). \end{align} $$
Performing repeated integration yields $$ \begin{align} \operatorname{E}[X] &=\operatorname{E}_Y[\operatorname{E}_X[X\mid Y]]\tag{1} \end{align} $$ Applying $(1)$ to $X^2$ and using the fact that $\operatorname{Var}[X]=\operatorname{E}\left[X^2\right]-\operatorname{E}[X]^2$, we get $$ \begin{align} \operatorname{E}\left[X^2\right] &=\operatorname{E}_Y\left[\operatorname{E}_X\left[X^2\mid Y\right]\right]\\ &=\operatorname{E}_Y\left[\operatorname{Var}_X[X\mid Y]\right] +\operatorname{E}_Y\left[\operatorname{E}_X[X\mid Y]^2\right]\tag{2} \end{align} $$ Applying $(1)$ and $(2)$, we get $$ \begin{align} \operatorname{Var}[X] &=\operatorname{E}\left[X^2\right]-\operatorname{E}[X]^2\\ &=\operatorname{E}_Y\left[\operatorname{Var}_X[X\mid Y]\right] + \operatorname{E}_Y\left[\operatorname{E}_X[X\mid Y]^2\right]-\operatorname{E}_Y[\operatorname{E}_X[X|Y]]^2\\ &=\operatorname{E}_Y\left[\operatorname{Var}_X[X\mid Y] \right] +\operatorname{Var}_Y[\operatorname{E}_X[X\mid Y]]\tag{3} \end{align} $$ Now apply $(3)$ to the problem: $$ \begin{align} \operatorname{Var}\left[\sum_{i=1}^NX_i\right] &=\operatorname{E}_N\left[\left.\operatorname{Var}_X\left[\sum_{i=1}^NX_i\right]\right|N\right]+\operatorname{Var}_N\left[\left.\operatorname{E}_X\left[\sum_{i=1}^N X_i\right]\right|N\right]\\ &=\operatorname{E}_N[N\operatorname{Var}[X]]+\operatorname{Var}_N[N\operatorname{E}[X]]\\ &=\operatorname{E}[N]\operatorname{Var}[X]+\operatorname{Var}[N]\operatorname{E}[X]^2\tag{4} \end{align} $$
Another way to do this: let $Y_i = 1$ if $N \ge i$, $0$ otherwise. Then your sum is $$S = \sum_{i=1}^N X_i = \sum_{i=1}^\infty Y_i X_i$$ (I won't worry about convergence of infinite sums: if you wish you can use a truncated version of $N$ and then take limits). So $$\text{var}(S) = \sum_{i=1}^\infty \text{var}(Y_i X_i) + 2 \sum_{i=1}^\infty \sum_{j=1}^{i-1} \text{cov}(Y_i X_i, Y_j X_j)$$ Now $\text{var}(Y_i X_i) = E[Y_i^2 X_i^2] - E[Y_i X_i]^2 = E[Y_i] \text{var}(X) + \text{var}(Y_i) E[X]^2$, while for $j < i$, $\text{cov}(Y_i X_i, Y_j X_j) = E[Y_i Y_j X_i X_j] - E[Y_i X_i] E[Y_j X_j] = E[Y_i] (1 - E[Y_j]) E[X]^2$, so that $$ \text{var}(S) = \sum_{i=1}^\infty E[Y_i] \text{var}(X) + \sum_{i=1}^\infty \text{var}(Y_i) E[X]^2 + 2 \sum_{i=1}^\infty \sum_{j=1}^{i-1} E[Y_i](1 - E[Y_j]) E[X]^2$$ Now doing the same calculation with $X_i$ replaced by 1 (since $N = \sum_{i=1}^\infty Y_i$), $$ \text{var}(N) = \sum_{i=1}^\infty \text{var}(Y_i) + 2 \sum_{i=1}^\infty \sum_{j=1}^{i-1} E[Y_i](1 - E[Y_j])$$ so that $$ \text{var}(S) = \sum_{i=1}^\infty E[Y_i] \text{var}(X) + \text{var}(N) E[X]^2 = E[N] \text{var}(X) + \text{var}(N) E[X]^2 $$