Lagrangian mechanics, the traditional Newtonian mechanics you're probably used to, and Hamiltonian mechanics (the third paradigm for classical mechanics) are all equivalent and differ more or less only in how they give you the differential equations of motion.
Let $S$ be an abbreviation for the state of a system, like the set of all relevant positions and velocities, and the current time. In Newtonian mechanics, you have $\vec{F} = m\vec{a}$ for each particle/object. I'll rewrite this as
$$ \ddot{\vec{x}}_i = \frac{1}{m_i} \vec{F}_i(S). $$
That's your (set of) differential equation(s), and once you have the form(s) of $\vec{F}_i$, you can use whatever mathematical techniques you have at hand to solve. Only then do you plug in initial conditions to select which specific solution from your family of solutions to use.
By definition of "free particle," there are no forces, so your one object has $\ddot{\vec{x}} = 0$, which is easily solved to get $\vec{x} = \vec{a} + \vec{b}t$. It's pretty easy to see then that $\vec{a}$ has components $x(0)$ and $y(0)$, and $\vec{b}$ has components $v_x(0)$ and $v_y(0)$.
Here's how Lagrangian mechanics does the same thing. We define a new scalar function $L$ of the state $S$, and it is the kinetic energy minus any potential energy that might be around. (Note: this is not the action $S$ - I'm just using a nonstandard abbreviation.) Our differential equations are now
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\partial L(S)}{\partial \dot{q}_k}\right) - \frac{\partial L(S)}{\partial q_k} = 0. $$
(Here I'm taking $S$ to be a function of generalized coordinates $q_k$, which can be anything from positions to angles to whatever. Usually there are fewer $q_k$'s than there are components of all the $\vec{x}_i$'s - at least there are whenever there are constraints on the system - and this is part of the reason Lagrangian mechanics makes some problems easier.)
In the single free particle case we have no potential energy and so $L(S) = (1/2) m (\dot{x}^2 + \dot{y}^2)$. Treating $x$, $y$, $\dot{x}$, $\dot{y}$, and $t$ as independent variables, we see
$$ \frac{\partial L(S)}{\partial x} = \frac{\partial L(S)}{\partial y} = 0, \\
\frac{\partial L(S)}{\partial \dot{x}} = m \dot{x}, \\
\frac{\partial L(S)}{\partial \dot{y}} = m \dot{y}. $$
Differentiating with respect to $t$ (where $x$, $y$, $\dot{x}$, and $\dot{y}$ are now considered functions of time), we find our two equations
$$ m \ddot{x} = m \ddot{y} = 0. $$
Of course you can divide through by $m$ and get the exact same equations we had in the Newtonian case. This is because my $q_k$'s were the same as the components of my $\vec{x}_i$'s - there was no reason in this case to choose a different set of coordinates.
At this point you also have a differential equation, and that's it for the physics. You still have to use your differential equation skills to solve, and you still have to choose the correct constants of integration to reflect your particular initial conditions.
In summary: when you tried to jump right into the formula for the displacement at time $t$, you actually implicitly did all the $\vec{F} = m\vec{a}$ stuff in your head and skipped right to the plugging in of initial conditions. There is no "Lagrangian way" to plug in initial conditions; the methods only differ in how you get the equation(s) of motion.
