Your description of the Dirac delta ($\delta_0$) is incorrect. One of possible correct definitions is: $\delta_0$ is the derivative of the Heaviside function $h_0$, which is defined as $h_0(x)=1$ when $x\ge 0$; $h_0(x)=0$ when $x< 0$. The meaning of the derivative is not the classical one; instead, it is a stipulation that the certain integral identities hold: Fundamental theorem of Calculus and integration by parts. In particular, $\int_a^b \delta_0 =h_0(b)-h_0(a)=1$ whenever $a<0<b$.
Now back to counting initial conditions. If we have $u(0,x)=x^2$ for all $x\in\mathbb R$, is this one condition? Or maybe infinitely many, because there's a condition for every $x\in\mathbb R$? I guess we take the position that this is one condition: the distribution of matter represented by $u$ is given at time $t=0$, and that's all we have.
Another example: $u(0,x)=h_0(x)$ for all $x\in\mathbb R$. This is also one initial condition, we just have another function here instead of $x^2$. Since the function $h_0$ is piecewise defined, the condition $u(0,x)=h_0(x)$ amounts to
$$u(0,x)=0 \text{ when } x<0 \ \text{ and } \ u(0,x)=1 \text{ when } x\ge 0$$
But this split is just our way of writing $h_0$. It is of no consequence for the mathematical problem.
Similarly, the initial condition $u(0,\cdot) =\delta_0$ is also one condition, which may appear as two (or three) depending on how $\delta_0$ is presented in an application-oriented text.
This is a classical Diffusion PDE, if I have understood your question correctly.
In fact its solution was marking the birthday of Fourier technique. The straight forward solution is really standard literature since 190 years and you will find it >>> here for the so called Heat Equation as invented by Fourier. It would be quite redundant to copy/paste all hereto.
When applying the technique you particularly arrive to the point where the constants A, B, C need to be at final steps identified, there you should apply instead, your own boundary/initial conditions.
This should equip you best for the solution of the problem.
Best Answer
Consider the same equation on $(-\infty,\infty)\times[0,\infty)$ with initial value $u(x,0)=-A\,\delta(x+x_0)+A\,\delta(x-x_0)$. Since the initial value is odd, the solution will be odd and satisfy the boundary condition $u(0,t)=0$.